$course$/top/Default for MSII/Test #11 $course$/top/Default for MSII/Test #11/L78-Linear Equations: Slope and Y-Intercept L78- line-yInt-Slope-Equation Give the y-intercept, slope and equation of the line. var board = JXG.JSXGraph.initBoard(BOARDID,{ axis:true,grid:true, boundingbox: [-11,11,11,-11], showCopyright: false, showNavigation: true,resize:{enabled:true,throttle:200} }); var graph = board.create('functiongraph', [function(x){ return {m}*x+{b};}, -11,11],{firstArrow:true,lastArrow:true,strokeColor:'green'}); ]]> var board = JXG.JSXGraph.initBoard(BOARDID,{ axis:true,grid:true, boundingbox: [-11,11,11,-11], showCopyright: false, showNavigation: false }); var graph = board.create('functiongraph', [function(x){ return {m}*x+{b};}, -10,10],{firstArrow:true,lastArrow:true,strokeColor:'green'}); var yInt=board.create('point',[0,{b}],{size:3,fillOpacity:0,strokeColor:'orange',name:''}); var ansY=board.create('line',[[0,{b}],[0,{b}+{mNum}]],{strokeColor:'red',straightFirst:false,straightLast:false,firstArrow:false,lastArrow:{type:2,size:4}}); var ansX=board.create('line',[[0,{b}+{mNum}],[{mDen},{b}+{mNum}]],{strokeColor:'blue',straightFirst:false,straightLast:false,firstArrow:false,lastArrow:{type:2,size:4}}); ]]> 3.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:7}; n={-2,3}; bn={1:99999}; direction={0,1}; 0,"-","+"); disb=abs(b);]]> none 0 1 0 3 [b,0,b] _err == 0 1 1 The order pair of the y-intercept is ({_1},{_2})
]]> The y-intercept is {b}.
The ordered pair of the y-intercept is (0,{b}).

]]> 1 1 0 2 [mNum,mDen] _err == 0 1 1 The slope is {_0} {_1} ]]> The slope is $\frac{{mNum}}{{mDen}}$.

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

]]> 2 1 1000 1 equation _err == 0 1 1 (Use the division bar to show a fraction for example $\frac{2}{3}$ should be typed 2/3)
The equation of the line is: y={_0}

]]> The equation of the line is y=$\frac{{mNum}}{{mDen}}$x {signb} {disb}.

]]> L78- line-yInt-Slope-Equation Give the y-intercept, slope and equation of the line. var board = JXG.JSXGraph.initBoard(BOARDID,{ axis:true,grid:true, boundingbox: [-11,11,11,-11], showCopyright: false, showNavigation: true,resize:{enabled:true,throttle:200} }); var graph = board.create('functiongraph', [function(x){ return {m}*x+{b};}, -10,10],{firstArrow:true,lastArrow:true,strokeColor:'green'}); ]]> var board = JXG.JSXGraph.initBoard(BOARDID,{ axis:true,grid:true, boundingbox: [-11,11,11,-11], showCopyright: false, showNavigation: false }); var graph = board.create('functiongraph', [function(x){ return {m}*x+{b};}, -10,10],{firstArrow:true,lastArrow:true,strokeColor:'green'}); var yInt=board.create('point',[0,{b}],{size:3,fillOpacity:0,strokeColor:'orange',name:''}); var ansY=board.create('line',[[0,{b}],[0,{b}+{mNum}]],{strokeColor:'red',straightFirst:false,straightLast:false,firstArrow:false,lastArrow:{type:2,size:4}}); var ansX=board.create('line',[[0,{b}+{mNum}],[{mDen},{b}+{mNum}]],{strokeColor:'blue',straightFirst:false,straightLast:false,firstArrow:false,lastArrow:{type:2,size:4}}); ]]> 3.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:7}; n={-2,3}; bn={1:99999}; direction={0,1}; 0,"-","+"); disb=abs(b);]]> none 0 1 0 3 [b,0,b] _err == 0 1 1 The order pair of the y-intercept is ({_1},{_2})
]]> The y-intercept is {b}.
The ordered pair of the y-intercept is (0,{b}).

]]> 1 1 0 2 [mNum,mDen] _err == 0 1 1 The slope is {_0} {_1} ]]> The slope is $\frac{{mNum}}{{mDen}}$.

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

]]> 2 1 1000 1 equation _err == 0 1 1 (Use the division bar to show a fraction for example $\frac{2}{3}$ should be typed 2/3)
The equation of the line is: y={_0}

]]> The equation of the line is y=$\frac{{mNum}}{{mDen}}$x {signb} {disb}.

]]> L78-Name slope Give the slope of the linear equation. var board = JXG.JSXGraph.initBoard(BOARDID,{ axis:true,grid:true, boundingbox: [-11,11,11,-11], showCopyright: false, showNavigation: true,resize:{enabled:true,throttle:200} }); var graph = board.create('functiongraph', [function(x){ return {m}*x+{b};}, -10,10],{firstArrow:true,lastArrow:true,strokeColor:'green'}); ]]> The slope is $\frac{{mNum}}{{mDen}}$.

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

]]> ({num1a}{var1} + {num2a}) + ({num1b}{var1} + {num2b}) Given {num1a}{var1} + {num2a} + {num1b}{var1} + {num2b} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a} + {num2b} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}) Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}) Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+b)+(cx+d) ({num1a}{var1} + {num2a}) + ({num1b}{var1} + {num2b})

]]> ({num1a}{var1} + {num2a}) + ({num1b}{var1} + {num2b}) Given {num1a}{var1} + {num2a} + {num1b}{var1} + {num2b} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a} + {num2b} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}) Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}) Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+b)+(cy+d) ({num1a}{var1} + {num2a}) + ({num1b}{var3} + {num2b})

]]> ({num1a}{var1} + {num2a}) + ({num1b}{var3} + {num2b}) Given {num1a}{var1} + {num2a} + {num1b}{var3} + {num2b} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var3}+ {num2a} + {num2b} Use the commutative property of addition to write the like terms next to each other. {num1a}{var1} + {num1b}{var3}+ ({num2a} + {num2b}) Use the distributive property to group the coefficients of the like terms {num1a}{var1} + {num1b}{var3}+ ({=num2a+num2b}) Add the coefficients {num1a}{var1} + {num1b}{var3} {signV2} {absCoefVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+b)+(dy+ax) ({num1a}{var1} + {num2a}) + ({num2b} +{num1b}{var1})

]]> ({num1a}{var1} + {num2a}) + ({num2b}} + {num1b}{var1}) Given {num1a}{var1} + {num2a} + {num2b} +{num1b}{var1} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a} + {num2b} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}) Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}) Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); coefVar1=pick((num1ab==0)&&(num2ab==0),coefVar1,join("",0)); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+b)+cx ({num1a}{var1} + {num2a}) + {num1b}{var1}

]]> ({num1a}{var1} + {num2a}) + {num1b}{var1} Given {num1a}{var1} + {num2a} + {num1b}{var1} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ {num2a} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ {num2a} Add the coefficients {coefVar1}{ansVar1} {signV2} {=abs(num2a)}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+by)+(-ax+dy) ({num1a}{var1} + {num2a}{var2}) + ({num1b}{var1} + {num2b}{var2})

]]> ({num1a}{var1} + {num2a}{var2}) + ({num1b}{var1} + {num2b}{var2}) Given {num1a}{var1} + {num2a}{var2} + {num1b}{var1} + {num2b}{var2} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+by)+(cx+dy) ({num1a}{var1} + {num2a}{var2}) + ({num1b}{var1} + {num2b}{var2})

]]> ({num1a}{var1} + {num2a}{var2}) + ({num1b}{var1} + {num2b}{var2}) Given {num1a}{var1} + {num2a}{var2} + {num1b}{var1} + {num2b}{var2} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+by)+(cx-dy) ({num1a}{var1} {num2a}{var2}) + ({num1b}{var1} +{num2b}{var2})

]]> ({num1a}{var1} + {num2a}{var2}) + ({num1b}{var1} + {num2b}{var2}) Given {num1a}{var1} + {num2a}{var2} + {num1b}{var1} + {num2b}{var2} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+by)+(dy+ax) ({num1a}{var1} + {num2a}{var2}) + ({num2b}{var2} +{num1b}{var1})

]]> ({num1a}{var1} + {num2a}{var2}) + ({num2b}{var2} + {num1b}{var1}) Given {num1a}{var1} + {num2a}{var2} + {num2b}{var2} +{num1b}{var1} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); coefVar1=pick((num1ab==0)&&(num2ab==0),coefVar1,join("",0)); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+by)+(dy-ax) ({num1a}{var1} + {num2a}{var2}) + ({num2b}{var2} +{num1b}{var1})

]]> ({num1a}{var1} + {num2a}{var2}) + ({num2b}{var2} + {num1b}{var1}) Given {num1a}{var1} + {num2a}{var2} + {num2b}{var2} +{num1b}{var1} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); coefVar1=pick((num1ab==0)&&(num2ab==0),coefVar1,join("",0)); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-(ax+by+ez)+(cx+dy+fz) ({num1a}{var1} + {num2a}{var2} + {num3a}{var3}) + ({num1b}{var1} + {num2b}{var2} +{num3b}{var3})

]]> ({num1a}{var1} + {num2a}{var2} + {num3a}{var3}) + ({num1b}{var1} + {num2b}{var2} + {num3b}{var3}) Given {num1a}{var1} + {num2a}{var2} + {num3a}{var3} + {num1b}{var1} + {num2b}{var2} + {num3b}{var3}) The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} + {num3a}{var3} + {num3b}{var3}) Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} + ({num3a} + {num3b}){var3} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} + ({num3a} + {num3b}){var3} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2} {signV3} {absCoefVar3}{ansVar3}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,"","+"),num2a+num2b,"*",var2,pick(num3ab>0,"","+"),num3a+num3b,"*",var3); coefVar1=pick(num1ab==0,pick(num1ab==1,join("",num1ab),""),""); coefVar2=pick(abs(num2ab)==0,pick(abs(num2ab)==1,join("",abs(num2ab)),""),""); coefVar3=pick(abs(num3ab)==0,pick(abs(num3ab)==1,join("",abs(num3ab)),""),""); absCoefVar2=pick(num2ab==0,pick(abs(num2ab)==1,join("",abs(num2ab)),""),""); absCoefVar3=pick(num3ab==0,pick(abs(num3ab)==1,join("",abs(num3ab)),""),""); ansVar1=pick(num1ab==0,join("",var1),""); ansVar2=pick(num2ab==0,join("",var2),""); ansVar3=pick(num3ab==0,join("",var3),""); signV2=pick(num2ab>0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); signV3=pick(num3ab>0,pick(num3ab<0,"","-"),pick(num2ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-ax+by+-ax+dy {num1a}{var1} + {num2a}{var2} + {num1b}{var1} + {num2b}{var2}

]]> {num1a}{var1} + {num2a}{var2} + {num1b}{var1} + {num2b}{var2} Given {num1a}{var1} + {num2a}{var2} + {num1b}{var1} + {num2b}{var2} The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 L79-Add Binomials-ax+by+ez+cx+dy+fz {num1a}{var1} + {num2a}{var2} + {num3a}{var3} + {num1b}{var1} + {num2b}{var2} +{num3b}{var3}

]]> {num1a}{var1} + {num2a}{var2} + {num3a}{var3} + {num1b}{var1} + {num2b}{var2} + {num3b}{var3} Given {num1a}{var1} + {num2a}{var2} + {num3a}{var3} + {num1b}{var1} + {num2b}{var2} + {num3b}{var3}) The parentheses are not needed because the binomials are being added. {num1a}{var1} + {num1b}{var1}+ {num2a}{var2} + {num2b}{var2} + {num3a}{var3} + {num3b}{var3}) Use the commutative property of addition to write the like terms next to each other. ({num1a} + {num1b}){var1}+ ({num2a} + {num2b}){var2} + ({num3a} + {num3b}){var3} Use the distributive property to group the coefficients of the like terms ({=num1a+num1b}){var1}+ ({=num2a+num2b}){var2} + ({num3a} + {num3b}){var3} Add the coefficients {coefVar1}{ansVar1} {signV2} {absCoefVar2}{ansVar2} {signV3} {absCoefVar3}{ansVar3}
Combine double signs.
Coefficients equal to 1 do not need to be written.
Terms with a zero coefficient are equal to zero, therefore do not need to be written. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. =0,"","+"),num2a+num2b,"*",var2,pick(num3ab>=0,"","+"),num3a+num3b,"*",var3); coefVar1=pick(num1ab==0,pick(num1ab==1,join("",num1ab),""),""); coefVar2=pick(abs(num2ab)==0,pick(abs(num2ab)==1,join("",abs(num2ab)),""),""); coefVar3=pick(abs(num3ab)==0,pick(abs(num3ab)==1,join("",abs(num3ab)),""),""); absCoefVar2=pick(num2ab==0,pick(abs(num2ab)==1,join("",abs(num2ab)),""),""); absCoefVar3=pick(num3ab==0,pick(abs(num3ab)==1,join("",abs(num3ab)),""),""); ansVar1=pick(num1ab==0,join("",var1),""); ansVar2=pick(num2ab==0,join("",var2),""); ansVar3=pick(num3ab==0,join("",var3),""); signV2=pick(num2ab>0,pick(num2ab<0,"","-"),pick(num1ab==0,"+","")); signV3=pick(num3ab>0,pick(num3ab<0,"","-"),pick(num2ab==0,"+","")); a={-100:100}; b={-100:100}; c={-100:100}; d={-100:100}; e={-100:100}; f={-100:100}; g={-100:100}; h={-100:100}; k={-100:100}; m={-100:100}; n={-100:100}; p={-100:100}; r={-100:100}; s={-100:100}; t={-100:100}; v={-100:100}; w={-100:100}; x={-100:100}; y={-100:100}; z={-100:100};]]> abc 0 1 1000 1 ans _err == 0 1 1 $course$/top/Default for MSII/Test #11/L80- roman Numerals 1-50 L80- read Roman Numerals (1-20) What number is represented by the Roman Numerals {=rNum[randNum]}?

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

{FB}

{FBWords}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randNum={1:21}; 1 + 1 = 2","III= 3
1 + 1+ 1 = 3", "IV=4
1 less than 5 = 4
When a smaller number is to the left of the larger number subtract the value.","V=5","VI=6
5 + 1 = 6
When the smaller value is on the right of a larger value, add the amounts.","VII=7
5+2=7","VIII=8
5+3=8","IX=9
When a smaller number is the left of a larger value, subtract it from the larger number.
1 less than 10 is 9","X=10", "XI=11
10+1=11","XII=12
10+2=12","XIII=13
10+3=13", "XIV=14
10 added to (1 less than 5)
since the 1 is to the left of V, or 5, subtract 1 from 5. then add it to X (10)
10 + (1 less than 5)","XV=15
10+5=15", "XVI=16
10+5+1=16", "XVII=17
10+5+2=17", "XVIII=18
18+5+3=18", "XIX=19
10+(1 less than 10)
IX means one less than 10
10+9=19", "XX=20
10+10=20"]; FB=""; FBWords=""; x=20+randNum; FB=pick(x-50>=0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1);]]> abc 0 1 0 1 randNum _err == 0 1 1 L80- read Roman Numerals (21-40) What number is represented by the Roman Numerals {=rNum[randNum]}?

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

{FB}

{FBWords}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randNum={1:21}; 20+ 1 + 1 = 22","XXIII= 23
20+ 1 + 1+ 1 = 23", "XXIV=4
20+ 1 less than 5 = 24
When a smaller number is to the left of the larger number subtract the value.","XXV=25","XXVI=26
20+5 + 1 = 26
When the smaller value is on the right of a larger value, add the amounts.","XXVII=27
20+5+2=27","XXVIII=28
20+5+3=28","XXIX=29
When a smaller number is the left of a larger value, subtract it from the larger number.
1 less than 10 is 9","XXX=30", "XXXI=31
30+1=31","XXXII=32
30+2=12","XXXIII=33
30+3=13", "XXXIV=34
30 added to (1 less than 5)
since the 1 is to the left of V, or 5, subtract 1 from 5. then add it to XXX (30)
30 + (1 less than 5)","XXXV=35
30+5=15", "XXXVI=36
30+5+1=16", "XXXVII=37
30+5+2=37", "XXXVIII=38
38+5+3=38", "XXXIX=39
30+(1 less than 10)
IX means one less than 10
30+9=39", "XXXX=40"]; FB=""; FBWords=""; x=20+randNum; FB=pick(x-50>=0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1);]]> abc 0 1 0 1 randNum+20 _err == 0 1 1 L80- read Roman Numerals (41-60) What number is represented by the Roman Numerals {=rNum[randNum]}?

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

{FB}

{FBWords}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randNum={1:21}; =0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1);]]> abc 0 1 0 1 randNum+40 _err == 0 1 1 L80- write Roman Numerals (1,5,10,50,100,500,1000) Write each number in Roman Numerals. (Hint: Use uppercase letters.)

{=aNum[randOrder[0]]} {#A}	{=aNum[randOrder[1]]} {#B}	{=aNum[randOrder[2]]} {#C}
{=aNum[randOrder[3]]} {#D}	{=aNum[randOrder[4]]}{#E}	{=aNum[randOrder[5]]} {#F}
{=aNum[randOrder[6]]} {#G}

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

]]> 7.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randOrder=shuffle([0,1,2,3,4,5,6]); =0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1);]]> abc 0 #A 1 1000 1 rNum[randOrder[0]] _err == 0 1 1 1 #B 1 1000 1 rNum[randOrder[1]] _err == 0 1 1 2 #C 1 1000 1 rNum[randOrder[2]] _err == 0 1 1 3 #D 1 1000 1 rNum[randOrder[3]] _err == 0 1 1 4 #E 1 1000 1 rNum[randOrder[4]] _err == 0 1 1 5 #F 1 1000 1 rNum[randOrder[5]] _err == 0 1 1 6 #G 1 1000 1 rNum[randOrder[6]] _err == 0 1 1 L80- write Roman Numerals (1-20) Write {randNum} in Roman Numerals.

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

{FB}

{FBWords}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randNum={1:21}; 1 + 1 = 2","III= 3
1 + 1+ 1 = 3", "IV=4
1 less than 5 = 4
When a smaller number is to the left of the larger number subtract the value.","V=5","VI=6
5 + 1 = 6
When the smaller value is on the right of a larger value, add the amounts.","VII=7
5+2=7","VIII=8
5+3=8","IX=9
When a smaller number is the left of a larger value, subtract it from the larger number.
1 less than 10 is 9","X=10", "XI=11
10+1=11","XII=12
10+2=12","XIII=13
10+3=13", "XIV=14
10 added to (1 less than 5)
since the 1 is to the left of V, or 5, subtract 1 from 5. then add it to X (10)
10 + (1 less than 5)","XV=15
10+5=15", "XVI=16
10+5+1=16", "XVII=17
10+5+2=17", "XVIII=18
18+5+3=18", "XIX=19
10+(1 less than 10)
IX means one less than 10
10+9=19", "XX=20
10+10=20"]; I={2:1000}; II={2:1000}; III={3:1000}; IV={2:1000}; V={2:1000}; VI={3:1000}; VII={2:1000}; VIII={2:1000}; IX={3:1000}; X={2:1000}; XI={3:1000}; XII={2:1000}; XIII={2:1000}; XIV={2:1000}; XV={2:1000}; XVI={3:1000}; XVII={2:1000}; XVIII={2:1000}; XIX={2:1000}; XX={2:1000}; FB=""; FBWords=""; x=20+randNum; FB=pick(x-50>=0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1);]]> abc 0 1 1000 1 rNum[randNum] _err == 0 1 1 L80- write Roman Numerals (41-60) Write {=40+randNum} in Roman Numerals.

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

{FB}

{FBWords}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randNum={0:20}; =0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1); XL={2:1000}; XLI={2:1000}; XLII={2:1000}; XLIII={3:1000}; XLIV={2:1000}; XLV={2:1000}; XLVI={3:1000}; XLVII={2:1000}; XLVIII={2:1000}; XLIX={3:1000}; L={2:1000}; LI={3:1000}; LII={2:1000}; LIII={2:1000}; LIV={2:1000}; LV={2:1000}; LVI={3:1000}; LVII={2:1000}; LVIII={2:1000}; LIX={2:1000}; LX={2:1000};]]> abc 0 1 1000 1 rNum[randNum] _err == 0 1 1 L80- writeRoman Numerals (21-40) Write {=20+randNum} as a Roman Numeral.

]]> Roman Numerals I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000

{FB}

{FBWords}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. randNum={1:21}; 20+ 1 + 1 = 22","XXIII= 23
20+ 1 + 1+ 1 = 23", "XXIV=4
20+ 1 less than 5 = 24
When a smaller number is to the left of the larger number subtract the value.","XXV=25","XXVI=26
20+5 + 1 = 26
When the smaller value is on the right of a larger value, add the amounts.","XXVII=27
20+5+2=27","XXVIII=28
20+5+3=28","XXIX=29
When a smaller number is the left of a larger value, subtract it from the larger number.
1 less than 10 is 9","XXX=30", "XXXI=31
30+1=31","XXXII=32
30+2=12","XXXIII=33
30+3=13", "XXXIV=34
30 added to (1 less than 5)
since the 1 is to the left of V, or 5, subtract 1 from 5. then add it to XXX (30)
30 + (1 less than 5)","XXXV=35
30+5=15", "XXXVI=36
30+5+1=16", "XXXVII=37
30+5+2=37", "XXXVIII=38
38+5+3=38", "XXXIX=39
30+(1 less than 10)
IX means one less than 10
30+9=39", "XXXX=40"]; XXI={2:1000}; XXII={2:1000}; XXIII={3:1000}; XXIV={2:1000}; XXV={2:1000}; XXVI={3:1000}; XXVII={2:1000}; XXVIII={2:1000}; XXIX={3:1000}; XXX={2:1000}; XXXI={3:1000}; XXXII={2:1000}; XXXIII={2:1000}; XXXIV={2:1000}; XXXV={2:1000}; XXXVI={3:1000}; XXXVII={2:1000}; XXXVIII={2:1000}; XXXIX={2:1000}; XXXX={2:1000}; FB=""; FBWords=""; x=20+randNum; FB=pick(x-50>=0,FB,join("",FB,"L")); FBWords=pick(x-50>=0,FBWords,join("",FBWords,"50 is written with the symbol L
")); x=pick(x-50>=0,x,x-50); FB=pick(x-40>=0,FB,join("",FB,"XL")); FBWords=pick(x-40>=0,FBWords,join("",FBWords,"40 is written as ten less than 50, write XL
")); x=pick(x-40>=0,x,x-40); FB=pick(x-30>=0,FB,join("",FB,"XXX")); FBWords=pick(x-30>=0,FBWords,join("",FBWords,"30 is written as three tens, write XXX
")); x=pick(x-30>=0,x,x-30); FB=pick(x-20>=0,FB,join("",FB,"XX")); FBWords=pick(x-20>=0,FBWords,join("",FBWords,"20 is written as two tens, write XX
")); x=pick(x-20>=0,x,x-20); FB=pick(x-10>=0,FB,join("",FB,"X")); FBWords=pick(x-10>=0,FBWords,join("",FBWords,"10 is written with the symbol X
")); x=pick(x-10>=0,x,x-10); FB=pick(x-9>=0,FB,join("",FB,"IX")); FBWords=pick(x-9>=0,FBWords,join("",FBWords,"9 is written as one less than ten, write IX
")); x=pick(x-9>=0,x,x-9); FB=pick(x-8>=0,FB,join("",FB,"VIII")); FBWords=pick(x-8>=0,FBWords,join("",FBWords,"8 is written as three more than 5 VIII
")); x=pick(x-8>=0,x,x-8); FB=pick(x-7>=0,FB,join("",FB,"VII")); FBWords=pick(x-7>=0,FBWords,join("",FBWords,"7 is written as two more than 5 VII
")); x=pick(x-7>=0,x,x-7); FB=pick(x-6>=0,FB,join("",FB,"VI")); FBWords=pick(x-6>=0,FBWords,join("",FBWords,"6 is written as one more than 5 VI
")); x=pick(x-6>=0,x,x-6); FB=pick(x-5>=0,FB,join("",FB,"V")); FBWords=pick(x-5>=0,FBWords,join("",FBWords,"5 is written with the symbol V
")); x=pick(x-5>=0,x,x-5); FB=pick(x-4>=0,FB,join("",FB,"IV")); FBWords=pick(x-4>=0,FBWords,join("",FBWords,"4 is written as 1 less than 5 IV
")); x=pick(x-4>=0,x,x-4); FB=pick(x-3>=0,FB,join("",FB,"III")); FBWords=pick(x-3>=0,FBWords,join("",FBWords,"3 is written as three ones III
")); x=pick(x-3>=0,x,x-3); FB=pick(x-2>=0,FB,join("",FB,"II")); FBWords=pick(x-2>=0,FBWords,join("",FBWords,"2 is written as two ones II
")); x=pick(x-2>=0,x,x-2); FB=pick(x-1>=0,FB,join("",FB,"I")); FBWords=pick(x-1>=0,FBWords,join("",FBWords,"1 is written with the symbol I
")); x=pick(x-1>=0,x,x-1);]]> abc 0 1 1000 1 rNum[randNum] _err == 0 1 1 $course$/top/Default for MSII/Test #11/L81- Distributive Property and Signed Numbers L81- Distribute sign Number +a(-V+c) Simplify: {a}(-{var} + {c})

]]> {a}({b}{var} + {c}) Given

({a})({b}{var}) + ({a})({c})
Distribute {a} to both terms of the binomial {b}{var}+{c}

{=a*b}{var} + {=a*c}
({a})(-1){var} = {=-1*a}{var}
({a})({c}) = {=a*c}

{=a*b}{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Distribute sign Number -1(-V+c) Simplify: -(-{var} + {c})

]]> -(-{var} + {c}) Given

(-1)(-{var}) + (-1)({c}) A negative sign in front of a parentheses means -1 times the amount in the parentheses.
Distribute -1 to both terms of the binomial -{var}+{c}

{=a*b}{var} + {=a*c} Multiplying by -1 means change the sign of each term.
(-1)(-{var}) = {=-1*b}{var} = {var}
(-1)({c}) = -1{c}

{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
When the coefficient is 1 it does not need to be written. 1{var} = {var}
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Distribute sign Number -1(bV+c) Simplify: -({b}{var} + {c})

]]> -({b}{var} + {c}) Given

(-1)({b}{var}) + (-1)({c}) A negative sign in front of a parentheses means -1 times the amount in the parentheses.
Distribute -1 to both terms of the binomial {b}{var}+{c}

{=a*b}{var} + {=a*c} Multiplying by -1 means change the sign of each term.
(-1){b}{var} = {=-1*b}{var}
(-1)({c}) = -1{c}

{=a*b}{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
Combine the signs (+)(-) = (-)
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Distribute sign Number -a(+bv -c) Simplify: {a}({b}{var} {c})

]]> {a}({b}{var} + {c}) Given

({a})({b}{var}) + ({a})({c}) Distribute {a} to both terms of the binomial {b}{var}+{c}
Notice the negative sign is distributed with {=abs(a)} because it's value is {a}.

{=a*b}{var} + {=a*c} Multiply ({a})({b}) = {=a*b}
Multiply ({a})({c}) = {=a*c}, write {signAC} {=abs(a*c)}

{=a*b}{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Distribute sign Number -a(-bV+c) Simplify: {a}({b}{var} + {c})

]]> {a}({b}{var} + {c}) Given

({a})({b}{var}) + ({a})({c}) Distribute {a} to both terms of the binomial {b}{var}+{c}
Notice the negative sign is distributed with {=abs(a)} because it's value is {a}.

{=a*b}{var} + {=a*c} Multiply ({a})({b}) = {=a*b}
Multiply ({a})({c}) = {=a*c}

{=a*b}{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
Combine the signs (+)(-) = (-)
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Distribute sign Number -a(bV+c) Simplify: {a}({b}{var} + {c})

]]> {a}({b}{var} + {c}) Given

({a})({b}{var}) + ({a})({c}) Distribute {a} to both terms of the binomial {b}{var}+{c}
Notice the negative sign is distributed with {=abs(a)} because it's value is {a}.

{=a*b}{var} + {=a*c} Multiply ({a})({b}) = {=a*b}
Multiply ({a})({c}) = {=a*c}

{=a*b}{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
Combine the signs (+)(-) = (-)
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Signed - Term plus Distribute sign Number a(bv +c) Simplify: {a}({b}{var} {c})

]]> {a}({b}{var} + {c}) Given

({a})({b}{var}) + ({a})({c}) Distribute {a} to both terms of the binomial {b}{var}+{c}
Notice the negative sign is distributed with {=abs(a)} because it's value is {a}.

{=a*b}{var} + {=a*c} Multiply ({a})({b}) = {=a*b}
Multiply ({a})({c}) = {=a*c}, write {signAC} {=abs(a*c)}

{=a*b}{var} {signAC} {=abs(a*c)}
The terms are NOT like-terms.
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 $course$/top/Default for MSII/Test #11/L81- Distributive Property and Signed Numbers/L81- Algebraic Expressions with Distributive Property L81- Term plus distribute sign Number dV+a(bV+c) Simplify: {d}{var} {signA} {a}({b}{var} + {c})

]]> {d}{var} {signA} {a}({b}{var} + {c}) Given

{d}{var} {signA} ({a})({b}{var}) + ({a})({c}) Distribute {a} to both terms of the binomial {b}{var}+{c}
Notice the negative sign is distributed with {=abs(a)} because it's value is {a}.

{d}{var} {signA} {=a*b}{var} + {=a*c} Multiply ({a})({b}) = {=a*b}
Multiply ({a})({c}) = {=a*c}, write {signAC} {=abs(a*c)}

{=d+a*b}{var} {signAC} {=abs(a*c)}
The terms {d}{var} and {=a*b}{var} are like-terms.
{d}{var} + {=a*b}{var} = {=d+a*b}{var}
This is the final answer. ]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 L81- Term plus distribute sign Number dV-a(-bV+c) Simplify: {d}{var} {signA} {a}({b}{var} + {c})

]]> {d}{var} {signA} {a}({b}{var} + {c}) Given

{d}{var} {signA} ({a})({b}{var}) + ({a})({c})
Distribute {a} to both terms of the binomial {b}{var}+{c}

{d}{var} {signAB} {=a*b}{var} + {=a*c}
({a})({b}){var} = {=b*a}{var}
({a})({c}) = {=a*c}

{=d+a*b}{var} + {=(a*c)}

The terms {d}{var} and {=a*b}{var} are like-terms, combine them.
{d}{var} + {=a*b}{var} = {=d+a*b}{var}

{=d+a*b}{var} {signAC} {=abs(a*c)}

Combine double signs
(+)(-) = (-)

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 1000 1 ans _err == 0 1 1 $course$/top/Default for MSII/Test #11/L82 - Cylinders: Volume and Surface Area L82-Surface Area-cylinder (given diameter) Use the formula TSA=2$\pi$r²+$\pi$dh var board=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-5,8,5,-3], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board.create('point', [-3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board.create('point', [ 3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board.create('point', [0, -2],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p1, p2, p3, Math.PI,6.28] ); var p4 = board.create('point', [-3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board.create('point', [ 3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p6 = board.create('point', [0, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p4, p5, p6], {strokeColor:'green'}); var pr1=board.create('point',[3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2=board.create('point',[3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var pl1=board.create('point',[-3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pl2=board.create('point',[-3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var heightR=board.create('line',[pr1,pr2],{straightFirst:false,straightLast:false,strokeWidth:1}); var heightL=board.create('line',[pl1,pl2],{straightFirst:false,straightLast:false,strokeWidth:1}); var tRadius=board.create('text',[0.1,5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var tHeight=board.create('text',[3.7,2.5,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'20'}); var diameterL=board.create('line',[pr2,pl2],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); ]]> Total Surface Area (TSA) of the Cylinder means the sum of the area of each surface

TSA = 2$\pi$r² + $\pi$dh		Total Surface Area = The area of the two bases plus the area of the rectangle between them.
TSA = 2(3.14)({radius})² + (3.14)({diameter})({height})		Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter} Calculate the radius. Radius = $\frac{1}{2}$ of the diameter = ($\frac{1}{2}$)({diameter})={radius} height = {height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		{radius}² = {radius}{radius} = {=radiusradius} ({diameter})({height}) = {=diameter*height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		2(3.14)({=radiusradius}) = {=23.14radiusradius}
TSA = {=23.14radiusradius} + {=3.14diameter*height}		(3.14)({=diameterheight}) = {=3.14diameter*height}
TSA = {TSA} {unit}²		Add {=23.14radiusradius} and {=3.14diameterheight} {=23.14radiusradius} + {=3.14diameterheight} = {TSA}

The formula will calculate the total surface area, however, if you want to understand the process and how the formula is derived, a step-by-step approach to finding the Total Surface Area is below.

Area of the top and bottom bases
var board1=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-6,6,6,-6], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p4 = board1.create('point', [0,0],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board1.create('point', [0, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var base = board1.create('circle', [ p4, p5], {strokeColor:'green',fillColor:'pink'}); var tRadius=board1.create('text',[0,0.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var diameterL=board1.create('line',[[-5,0],[5,0]],{fixed:true,straightFirst:false,straightLast:false,dash:1,strokeColor:'red'});	Area = $\pi$ radius²	The Formula for Area
	Area = (3.14)({radius})²	Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter}, therefore calculate the radius. Radius = $\frac{1}{2}$ of the diameter = ($\frac{1}{2}$)({diameter})={radius}
	Area = (3.14)({=radius*radius})	Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
	Area = {area} {unit}²	Multiply (3.14)({=radius*radius})
		*** Remember that Area is square units

Area of the rectangle (the section between the bases)
var board2=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-6,6,6,-6], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board2.create('point', [-3,4],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board2.create('point', [3, 4],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board2.create('point', [3,-4],{fixed:true,size:'0',name:'',showInfobox:false}); var p4 = board2.create('point', [-3,-4],{fixed:true,size:'0',name:'', showInfobox:false}); var base = board2.create('polygon', [ p1, p2,p3,p4], {strokeColor:'black',fillColor:'blue'}); var tCircumference=board2.create('text',[0,4.1,"Circumference of the circle"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'15'}); var tHeight=board2.create('text',[4.1,0,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'15'});	Area = lw Area = $\pi$dh	Formula for Area of a rectangle The length is the circumference of the circle (c=$\pi$d) The width is the height of the cylinder (h)
	Area = (3.14)({diameter})({height})	Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter} h is height = {height} {unit}
	Area= (3.14)({=diameter*height})	Calculate ({diameter})({height}) ({diameter})({height}) = {=diameter*height}
	*Area= {=cheight} {unit}²**	Multiply (3.14)({=diameter*height})

Area of the top and bottom bases + Area of the rectangle (the section between the bases)
var board3=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-8,8,8,-8], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p4 = board3.create('point', [0,5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board3.create('point', [0, 7],{fixed:true,size:'0',name:'',showInfobox:false}); var base = board3.create('circle', [ p4, p5], {strokeColor:'green',fillColor:'pink'}); var tRadius=board3.create('text',[0,5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'5'}); var diameterL=board3.create('line',[[-2,5],[2,5]],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); var p1 = board3.create('point', [0,-5],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board3.create('point', [0, -7],{fixed:true,size:'0',name:'',showInfobox:false}); var base2 = board3.create('circle', [ p1, p2], {strokeColor:'green',fillColor:'pink'}); var tRadiusB2=board3.create('text',[0,-5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'5'}); var diameterB2L=board3.create('line',[[-2,-5],[2,-5]],{fixed:true,straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); var pr1 = board3.create('point', [-4,3],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2 = board3.create('point', [4, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var pr3 = board3.create('point', [4,-3],{fixed:true,size:'0',name:'',showInfobox:false}); var pr4 = board3.create('point', [-4,-3],{fixed:true,size:'0',name:'', showInfobox:false}); var base = board3.create('polygon', [ pr1, pr2,pr3,pr4], {strokeColor:'black',fillColor:'blue'}); var tCircumference=board3.create('text',[0,2.9,"Circumference of the circle"],{fixed:true,anchorX:'middle',anchorY:'top',fontSize:'5'}); var tHeight=board3.create('text',[4.05,0,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'5'});	Area of top base = {area} {unit}²Area of bottom base= {area} {unit}²*Area of rectangle connecting bases = {=cheight} {unit}²**	The area of each section
	TSA= {area} + {area} + {=c*height}	Total Surface area is the sum of the areas
	Area= {TSA} {unit}²

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. 2"),join("",unit,"³")]; Xa=cos(angle); Ya=sin(angle); cir=3.14*2*radius; area=3.14*radius*radius; supAngle=angle+pi(); Xb=cos(supAngle); Yb=sin(supAngle); volume=area*height; c=3.14*diameter; TSA=2*area+c*height;]]> none 0 1 0 2 [TSA,unitNum] crit1+crit2 1 1 Find the total surface area of the cylinder. (use 3.14 for π)
Round to 2 decimal places

{_0}	{_1:units}

]]> L82-Surface Area-cylinder (given diameter) Use the formula TSA=2$\pi$r²+$\pi$dh var board=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-5,8,5,-3], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board.create('point', [-3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board.create('point', [ 3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board.create('point', [0, -2],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p1, p2, p3, Math.PI,6.28] ); var p4 = board.create('point', [-3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board.create('point', [ 3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p6 = board.create('point', [0, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p4, p5, p6], {strokeColor:'green'}); var pr1=board.create('point',[3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2=board.create('point',[3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var pl1=board.create('point',[-3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pl2=board.create('point',[-3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var heightR=board.create('line',[pr1,pr2],{straightFirst:false,straightLast:false,strokeWidth:1}); var heightL=board.create('line',[pl1,pl2],{straightFirst:false,straightLast:false,strokeWidth:1}); var tRadius=board.create('text',[0.1,5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var tHeight=board.create('text',[3.7,2.5,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'20'}); var diameterL=board.create('line',[pr2,pl2],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); ]]> Total Surface Area (TSA) of the Cylinder means the sum of the area of each surface

TSA = 2$\pi$r² + $\pi$dh		Total Surface Area = The area of the two bases plus the area of the rectangle between them.
TSA = 2(3.14)({radius})² + (3.14)({diameter})({height})		Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter} Calculate the radius. Radius = $\frac{1}{2}$ of the diameter = ($\frac{1}{2}$)({diameter})={radius} height = {height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		{radius}² = {radius}{radius} = {=radiusradius} ({diameter})({height}) = {=diameter*height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		2(3.14)({=radiusradius}) = {=23.14radiusradius}
TSA = {=23.14radiusradius} + {=3.14diameter*height}		(3.14)({=diameterheight}) = {=3.14diameter*height}
TSA = {TSA} {unit}²		Add {=23.14radiusradius} and {=3.14diameterheight} {=23.14radiusradius} + {=3.14diameterheight} = {TSA}

Area of the top and bottom bases
var board1=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-6,6,6,-6], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p4 = board1.create('point', [0,0],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board1.create('point', [0, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var base = board1.create('circle', [ p4, p5], {strokeColor:'green',fillColor:'pink'}); var tRadius=board1.create('text',[0,0.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var diameterL=board1.create('line',[[-5,0],[5,0]],{fixed:true,straightFirst:false,straightLast:false,dash:1,strokeColor:'red'});	Area = $\pi$ radius²	The Formula for Area
	Area = (3.14)({radius})²	Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter}, therefore calculate the radius. Radius = $\frac{1}{2}$ of the diameter = ($\frac{1}{2}$)({diameter})={radius}
	Area = (3.14)({=radius*radius})	Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
	Area = {area} {unit}²	Multiply (3.14)({=radius*radius})
		*** Remember that Area is square units

Area of the rectangle (the section between the bases)
var board2=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-6,6,6,-6], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board2.create('point', [-3,4],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board2.create('point', [3, 4],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board2.create('point', [3,-4],{fixed:true,size:'0',name:'',showInfobox:false}); var p4 = board2.create('point', [-3,-4],{fixed:true,size:'0',name:'', showInfobox:false}); var base = board2.create('polygon', [ p1, p2,p3,p4], {strokeColor:'black',fillColor:'blue'}); var tCircumference=board2.create('text',[0,4.1,"Circumference of the circle"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'15'}); var tHeight=board2.create('text',[4.1,0,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'15'});	Area = lw Area = $\pi$dh	Formula for Area of a rectangle The length is the circumference of the circle (c=$\pi$d) The width is the height of the cylinder (h)
	Area = (3.14)({diameter})({height})	Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter} h is height = {height} {unit}
	Area= (3.14)({=diameter*height})	Calculate ({diameter})({height}) ({diameter})({height}) = {=diameter*height}
	*Area= {=cheight} {unit}²**	Multiply (3.14)({=diameter*height})

Area of the top and bottom bases + Area of the rectangle (the section between the bases)
var board3=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-8,8,8,-8], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p4 = board3.create('point', [0,5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board3.create('point', [0, 7],{fixed:true,size:'0',name:'',showInfobox:false}); var base = board3.create('circle', [ p4, p5], {strokeColor:'green',fillColor:'pink'}); var tRadius=board3.create('text',[0,5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'5'}); var diameterL=board3.create('line',[[-2,5],[2,5]],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); var p1 = board3.create('point', [0,-5],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board3.create('point', [0, -7],{fixed:true,size:'0',name:'',showInfobox:false}); var base2 = board3.create('circle', [ p1, p2], {strokeColor:'green',fillColor:'pink'}); var tRadiusB2=board3.create('text',[0,-5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'5'}); var diameterB2L=board3.create('line',[[-2,-5],[2,-5]],{fixed:true,straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); var pr1 = board3.create('point', [-4,3],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2 = board3.create('point', [4, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var pr3 = board3.create('point', [4,-3],{fixed:true,size:'0',name:'',showInfobox:false}); var pr4 = board3.create('point', [-4,-3],{fixed:true,size:'0',name:'', showInfobox:false}); var base = board3.create('polygon', [ pr1, pr2,pr3,pr4], {strokeColor:'black',fillColor:'blue'}); var tCircumference=board3.create('text',[0,2.9,"Circumference of the circle"],{fixed:true,anchorX:'middle',anchorY:'top',fontSize:'5'}); var tHeight=board3.create('text',[4.05,0,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'5'});	Area of top base = {area} {unit}²Area of bottom base= {area} {unit}²*Area of rectangle connecting bases = {=cheight} {unit}²**	The area of each section
	TSA= {area} + {area} + {=c*height}	Total Surface area is the sum of the areas
	Area= {TSA} {unit}²

{_0}	{_1:units}

]]> L82-Volume-cylinder (given Diameter) Use the formula V=$\pi$r²h var board=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-5,8,5,-3], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board.create('point', [-3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board.create('point', [ 3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board.create('point', [0, -2],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p1, p2, p3, Math.PI,6.28] ); var p4 = board.create('point', [-3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board.create('point', [ 3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p6 = board.create('point', [0, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p4, p5, p6], {strokeColor:'green'}); var pr1=board.create('point',[3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2=board.create('point',[3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var pl1=board.create('point',[-3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pl2=board.create('point',[-3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var heightR=board.create('line',[pr1,pr2],{straightFirst:false,straightLast:false,strokeWidth:1}); var heightL=board.create('line',[pl1,pl2],{straightFirst:false,straightLast:false,strokeWidth:1}); var tRadius=board.create('text',[0.1,5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var tHeight=board.create('text',[3.7,2.5,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'20'}); var diameterL=board.create('line',[pr2,pl2],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); ]]> Area of the base times height
V = $\pi$r²h

Volume of the Cylinder
V = $\pi$r²h		Formula for volume
Volume = (3.14)({radius})²{height}		Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter}, therefore calculate the radius. Radius = $\frac{1}{2}$ of the diameter = ($\frac{1}{2}$)({diameter})={radius} h is height = {height} {unit}
Volume = (3.14)({=radius*radius})({height})		Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
*Volume = (3.14)({=radiusradiusheight})*		Multiply ({=radius*radius}) times {height}
Volume = {volume} {unit}³		Multiply (3.14)({=radiusradiusheight})

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 2 [volume,2] crit1=(_0==round(volume,3))*.75; crit2=(_1==unitNum)*.25; crit1+crit2 1 1 Find the volume of the cylinder. (use 3.14 for π)
Round to three decimal places.

{_0}{_1:units:MCE}]]> L82-Volume-cylinder (given Diameter) var board=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-5,8,5,-3], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board.create('point', [-3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board.create('point', [ 3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board.create('point', [0, -2],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p1, p2, p3, Math.PI,6.28] ); var p4 = board.create('point', [-3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board.create('point', [ 3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p6 = board.create('point', [0, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p4, p5, p6], {strokeColor:'green'}); var pr1=board.create('point',[3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2=board.create('point',[3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var pl1=board.create('point',[-3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pl2=board.create('point',[-3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var heightR=board.create('line',[pr1,pr2],{straightFirst:false,straightLast:false,strokeWidth:1}); var heightL=board.create('line',[pl1,pl2],{straightFirst:false,straightLast:false,strokeWidth:1}); var tRadius=board.create('text',[0.1,5.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var tHeight=board.create('text',[3.7,2.5,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'20'}); var diameterL=board.create('line',[pr2,pl2],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); ]]> Area of the base times height
V = $\pi$r²h

Volume of the Cylinder
V = $\pi$r²h		Formula for volume
Volume = (3.14)({radius})²{height}		Substitute the known values: Use $\pi$ = 3.14 diameter = {diameter}, therefore calculate the radius. Radius = $\frac{1}{2}$ of the diameter = ($\frac{1}{2}$)({diameter})={radius} h is height = {height} {unit}
Volume = (3.14)({=radius*radius})({height})		Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
*Volume = (3.14)({=radiusradiusheight})*		Multiply ({=radius*radius}) times {height}
Volume = {volume} {unit}³		Multiply (3.14)({=radiusradiusheight})

{_0}{_1:units:MCE}]]> L82-Volume-cylinder (given Radius) Use the formula V=$\pi$r²h var board=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-5,8,5,-3], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board.create('point', [-3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board.create('point', [ 3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board.create('point', [0, -2],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p1, p2, p3, Math.PI,6.28] ); var p4 = board.create('point', [-3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board.create('point', [ 3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p6 = board.create('point', [0, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p4, p5, p6], {strokeColor:'green'}); var pr1=board.create('point',[3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2=board.create('point',[3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var pl1=board.create('point',[-3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pl2=board.create('point',[-3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var heightR=board.create('line',[pr1,pr2],{straightFirst:false,straightLast:false,strokeWidth:1}); var heightL=board.create('line',[pl1,pl2],{straightFirst:false,straightLast:false,strokeWidth:1}); var tRadius=board.create('text',[1.8,5.1,"{radius} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var tHeight=board.create('text',[3.7,2.5,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'20'}); var diameterL=board.create('line',[[0,5],pr2],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); ]]> Area of the base times height
V = $\pi$r²h

Volume of the Cylinder
V = $\pi$r²h		Formula for volume
Volume = (3.14)({radius})²{height}		Substitute the known values: Use $\pi$ = 3.14 Radius = ={radius} {unit} h is height = {height} {unit}
Volume = (3.14)({=radius*radius})({height})		Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
*Volume = (3.14)({=radiusradiusheight})*		Multiply ({=radius*radius}) times {height}
Volume = {volume} {unit}³		Multiply (3.14)({=radiusradiusheight})

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 2 [volume,unitNum] crit1=(_0==round(volume,3))*.75; crit2=(_1==unitNum)*.25; crit1+crit2 1 1 Find the volume of the cylinder. (use 3.14 for π)
Round to three decimal places.

{_0}{_1:units:MCE}]]> L82-Volume-cylinder (given Radius) var board=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-5,8,5,-3], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board.create('point', [-3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board.create('point', [ 3, 0],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board.create('point', [0, -2],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p1, p2, p3, Math.PI,6.28] ); var p4 = board.create('point', [-3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board.create('point', [ 3, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var p6 = board.create('point', [0, 3],{fixed:true,size:'0',name:'',showInfobox:false}); var ell = board.create('ellipse', [ p4, p5, p6], {strokeColor:'green'}); var pr1=board.create('point',[3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pr2=board.create('point',[3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var pl1=board.create('point',[-3.61,0],{fixed:true,size:'0',name:'',showInfobox:false}); var pl2=board.create('point',[-3.61,5],{fixed:true,size:'0',name:'',showInfobox:false}); var heightR=board.create('line',[pr1,pr2],{straightFirst:false,straightLast:false,strokeWidth:1}); var heightL=board.create('line',[pl1,pl2],{straightFirst:false,straightLast:false,strokeWidth:1}); var tRadius=board.create('text',[1.8,5.1,"{radius} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var tHeight=board.create('text',[3.7,2.5,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'20'}); var diameterL=board.create('line',[[0,5],pr2],{straightFirst:false,straightLast:false,dash:1,strokeColor:'red'}); ]]> Area of the base times height
V = $\pi$r²h

Volume of the Cylinder
V = $\pi$r²h		Formula for volume
Volume = (3.14)({radius})²{height}		Substitute the known values: Use $\pi$ = 3.14 Radius = ={radius} {unit} h is height = {height} {unit}
Volume = (3.14)({=radius*radius})({height})		Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
*Volume = (3.14)({=radiusradiusheight})*		Multiply ({=radius*radius}) times {height}
Volume = {volume} {unit}³		Multiply (3.14)({=radiusradiusheight})

{_0}{_1:units:MCE}]]> $course$/top/Default for MSII/Test #11/L83- Multiplying Monomials L83-Multiplying exponent Monomials coefficient equals 1 Simplify: {var1}^{exp1}•{var1}^{exp2}

]]> Expand, then multiply.

{var1}^{exp1}•{var1}^{exp2}		Given
{factor1}•{factor2}		Write the expression as a product of factors
{var1}^{exp12}		The base {var1} is multiplied {exp12} times

The bases are equal, add the exponents

{var1}^{exp1}•{var1}^{exp2}		Given
{var1}^{{exp1}+{exp2}}		Add the Exponents
{var1}^{exp12}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 1 ans _err == 0 1 1 {var1}^{_0}

]]> L83-Multiplying Monomial times Monomial Simplify: ({num1}{var1}^{exp1})( {num2}{var1}^{exp2})

]]> Expand, then multiply.

({num1}{var1}^{exp1})( {num2}{var1}^{exp2})		Given
({num1}{factor1})({num2}{factor2})		Write the expression as a product of factors
({num1})({num2}){var1}^{exp12}		The coefficients are multiplied
{=num1*num2}{var1}^{exp12}		The base {var1} is multiplied {exp12} times

The bases are equal, add the exponents

({num1}{var1}^{exp1})({num2}{var1}^{exp2})		Given
{=num1*num2}{var1}^{{exp1}+{exp2}}		Add the Exponents
({num1})({num2}){var1}^{{exp1}+{exp2}}		Multiply the Coefficients
{=num1*num2}{var1}^{exp12}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 2 [num1*num2,ans] _err == 0 1 1 {_0}{var1}^{_1}

]]> L83-Multiplying Monomial times Monomials Simplify: {num1}{var1}^{exp1}• {num2}{var1}^{exp2}

]]> Expand, then multiply.

{num1}{var1}^{exp1}• {num2}{var1}^{exp2}		Given
{num1}{factor1}•{num2}{factor2}		Write the expression as a product of factors
({num1})({num2}){var1}^{exp12}		The coefficients are multiplied
{=num1*num2}{var1}^{exp12}		The base {var1} is multiplied {exp12} times

The bases are equal, add the exponents

{num1}{var1}^{exp1}• {num2}{var1}^{exp2}		Given
{=num1*num2}{var1}^{{exp1}+{exp2}}		Add the Exponents
({num1})({num2}){var1}^{{exp1}+{exp2}}		Multiply the Coefficients
{=num1*num2}{var1}^{exp12}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 2 [num1*num2,ans] _err == 0 1 1 {_0}{var1}^{_1}

]]> L83-Multiplying NEG Number Monomials Simplify: ({num1}{var1}^{exp1})( {num2}{var1}^{exp2})

]]> Expand, then multiply.

({num1}{var1}^{exp1})( {num2}{var1}^{exp2})		Given
({num1}{factor1})({num2}{factor2})		Write the expression as a product of factors
({num1})({num2}){var1}^{exp12}		The coefficients are multiplied
{=num1*num2}{var1}^{exp12}		The base {var1} is multiplied {exp12} times

The bases are equal, add the exponents

({num1}{var1}^{exp1})({num2}{var1}^{exp2})		Given
{=num1*num2}{var1}^{{exp1}+{exp2}}		Add the Exponents
({num1})({num2}){var1}^{{exp1}+{exp2}}		Multiply the Coefficients
{=num1*num2}{var1}^{exp12}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 2 [num1*num2,ans] _err == 0 1 1 {_0}{var1}^{_1}

]]> L83-Multiplying Number by Monomials Simplify: ({num1})( {num2}{var1}^{exp2})

]]> Expand, then multiply.

({num1})( {num2}{var1}^{exp2})		Given
({num1})({num2})({factor2})		Write the expression as a product of factors
({num1})({num2}){var1}^{exp2}		The coefficients are multiplied
{=num1*num2}{var1}^{exp2}		The base {var1} is multiplied {exp12} times

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. abc 0 1 0 2 [num1*num2,ans] _err == 0 1 1 {_0}{var1}^{_1}

]]> $course$/top/Default for MSII/Test #11/L84- Graph Line from equation L84 - Graph Equation y=mx+b Graph the equation $y={s} $
using the slope and y-intercept. {#A}]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. n={-2,-1,3}; m={-3:0,1:4}; bn={1:99999}; xVals=shuffle([-5,-4,-3,-2,-1,0,1,2,3,4,5]); 0,join("","Go down ",abs(m)," because the slope is negative"),join("","Go up ",abs(m)," because the slope is positive")); xVal=[xVals[0],xVals[1],xVals[2],xVals[3]]; xVal=sort(xVal); x1=xVal[0]; y1=m*x1+n; p1=join("",m,"(",x1,")",pick(n<0,join(""," + ",n),join("",n))); p1S1=join("",m*x1,pick(n<0,join(""," + ",n),join("",n))); p1S1FB=join("","Multiply ",m," and ",x1); p1S2=join("",m*x1+n); p1S2FB=join("",p1S1," = ",p1S2); x2=xVal[1]; y2=m*x2+n; p2=join("",m,"(",x2,")",pick(n<0,join(""," + ",n),join("",n))); p2S1=join("",m*x2,pick(n<0,join(""," + ",n),join("",n))); p2S1FB=join("","Multiply ",m," and ",x2); p2S2=join("",m*x2+n); p2S2FB=join("",p2S1," = ",p2S2); x3=xVal[2]; y3=m*x3+n; p3=join("",m,"(",x3,")",pick(n<0,join(""," + ",n),join("",n))); p3S1=join("",m*x3,pick(n<0,join(""," + ",n),join("",n))); p3S1FB=join("","Multiply ",m," and ",x3); p3S2=join("",m*x3+n); p3S2FB=join("",p3S1," = ",p3S2); x4=xVal[3]; y4=m*x4+n; p4=join("",m,"(",x4,")",pick(n<0,join(""," + ",n),join("",n))); p4S1=join("",m*x4,pick(n<0,join(""," + ",n),join("",n))); p4S1FB=join("","Multiply ",m," and ",x4); p4S2=join("",m*x4+n); p4S2FB=join("",p4S1," = ",p4S2);]]> none 0 #A 1 0 4 [0,n,1,m+n] ((_1==n+m*_0)*.5) + ((_3==n+m*_2)*.5) 1 1

]]> A line can be graphed from two points.

1. Locate the y-intercept. This is the first point.

2. Use the slope to find a second point.

The first point is the y-intercept.

Place the first point at (0,{n}) (The x-value of the y-intercept is always zero.)

The second point is calculated using the slope.

The slope is {m}. Write the slope as a fraction by writing {m} over 1.

slope = $\frac{Δy}{Δx}$ = $\frac{{m}}{1}$

First, Start at the y-intercept. Go {m} units on the y axis.

{mFB}

Next, go 1 unit on the x-axis. Put your second point at (1,{=m+n})

]]> L84 - Graph Equation y=mx+b (fractional slope)

Graph the linear equation y={signm}$\frac{{DmNum}}{{DmDen}}$x {signb}{disb} using the slope and y-intercept.

{#A}]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:-1,1:7}; bn={1:99999}; signM={-1,1}; 0,join("","Go down ",abs(mNum)," because the slope is negative"),join("","Go up ",abs(mNum)," because the slope is positive")); signb=pick(b>0,"-","+"); disb=abs(b); signm=pick(signM>0,"-",""); DmNum=abs(mNum); DmDen=abs(mDen); mNumPb=mNum+b;]]> none 0 #A 1 0 4 [0,b,mDen,mNum+b] ((_1==b+m*_0)*.5) + ((_3==b+m*_2)*.5) 1 1

]]>

1. Locate the y-intercept. This is the first point.

Place the first point at (0,{n}) (The x-value of the y-intercept is always zero.)

2. Use the slope to find a second point.

The slope is {signm}$\frac{{DmNum}}{{DmDen}}$.

First, Start at the y-intercept. Go {mNum} units on the y axis.

{mFB}

Next, go {mDen} on the x-axis. Put your second point at ({mDen},{mNumPb})

]]> L84 - Graph Equation y=mx+b (fractional slope) (y-int of 0)

Graph the linear equation y={signm}$\frac{{DmNum}}{{DmDen}}$x using the slope and y-intercept.

{#A}]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:-1,1:7}; bn={1:99999}; signM={-1,1}; 0,join("","Go down ",abs(mNum)," because the slope is negative"),join("","Go up ",abs(mNum)," because the slope is positive")); signb=pick(b>=0,"-","+"); disb=abs(b); signm=pick(signM>0,"-",""); DmNum=abs(mNum); DmDen=abs(mDen); mNumPb=mNum+b;]]> none 0 #A 1 0 4 [0,b,mDen,mNum+b] ((_1==b+m*_0)*.5) + ((_3==b+m*_2)*.5) 1 1

]]>

1. Locate the y-intercept. This is the first point.

Since 0 can be added to the equation without changing its value, the y-intercept = 0.

Place the first point at (0,{n}) (The x-value of the y-intercept is always zero.)

2. Use the slope to find a second point.

The slope is {signm}$\frac{{DmNum}}{{DmDen}}$.

First, Start at the y-intercept. Go {mNum} units on the y axis.

{mFB}

Next, go {mDen} on the x-axis. Put your second point at ({mDen},{mNumPb})

]]> L84 - Graph Equation y=mx+b (y-int of 0) Graph the equation $y={s} $
using the slope and y-intercept. {#A}]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. n={-2,-1,3}; m={-3:0,1:4}; bn={1:99999}; xVals=shuffle([-5,-4,-3,-2,-1,0,1,2,3,4,5]); 0,join("","Go down ",abs(m)," because the slope is negative"),join("","Go up ",abs(m)," because the slope is positive")); xVal=[xVals[0],xVals[1],xVals[2],xVals[3]]; xVal=sort(xVal); x1=xVal[0]; y1=m*x1+n; p1=join("",m,"(",x1,")",pick(n<0,join(""," + ",n),join("",n))); p1S1=join("",m*x1,pick(n<0,join(""," + ",n),join("",n))); p1S1FB=join("","Multiply ",m," and ",x1); p1S2=join("",m*x1+n); p1S2FB=join("",p1S1," = ",p1S2); x2=xVal[1]; y2=m*x2+n; p2=join("",m,"(",x2,")",pick(n<0,join(""," + ",n),join("",n))); p2S1=join("",m*x2,pick(n<0,join(""," + ",n),join("",n))); p2S1FB=join("","Multiply ",m," and ",x2); p2S2=join("",m*x2+n); p2S2FB=join("",p2S1," = ",p2S2); x3=xVal[2]; y3=m*x3+n; p3=join("",m,"(",x3,")",pick(n<0,join(""," + ",n),join("",n))); p3S1=join("",m*x3,pick(n<0,join(""," + ",n),join("",n))); p3S1FB=join("","Multiply ",m," and ",x3); p3S2=join("",m*x3+n); p3S2FB=join("",p3S1," = ",p3S2); x4=xVal[3]; y4=m*x4+n; p4=join("",m,"(",x4,")",pick(n<0,join(""," + ",n),join("",n))); p4S1=join("",m*x4,pick(n<0,join(""," + ",n),join("",n))); p4S1FB=join("","Multiply ",m," and ",x4); p4S2=join("",m*x4+n); p4S2FB=join("",p4S1," = ",p4S2);]]> none 0 #A 1 0 4 [0,n,1,m+n] ((_1==n+m*_0)*.5) + ((_3==n+m*_2)*.5) 1 1

]]> A line can be graphed from two points.

1. Locate the y-intercept. This is the first point.

2. Use the slope to find a second point.

The first point is the y-intercept.

Place the first point at (0,{n}) (The x-value of the y-intercept is always zero.)

The second point is calculated using the slope.

The slope is {m}. Write the slope as a fraction by writing {m} over 1.

slope = $\frac{Δy}{Δx}$ = $\frac{{m}}{1}$

First, Start at the y-intercept. Go {m} units on the y axis.

{mFB}

Next, go 1 unit on the x-axis. Put your second point at (1,{=m+n})

]]> L84- line-yInt-Slope-Equation Give the y-intercept, slope and equation of the line. y = $\frac{{mNum}}{{mDen}}$ {signb} {disb}]]> var board = JXG.JSXGraph.initBoard(BOARDID,{ axis:true,grid:true, boundingbox: [-11,11,11,-11], showCopyright: false, showNavigation: false }); var graph = board.create('functiongraph', [function(x){ return {m}*x+{b};}, -10,10],{firstArrow:true,lastArrow:true,strokeColor:'green'}); var yInt=board.create('point',[0,{b}],{size:3,fillOpacity:0,strokeColor:'orange',name:''}); var ansY=board.create('line',[[0,{b}],[0,{b}+{mNum}]],{strokeColor:'red',straightFirst:false,straightLast:false,firstArrow:false,lastArrow:{type:2,size:4}}); var ansX=board.create('line',[[0,{b}+{mNum}],[{mDen},{b}+{mNum}]],{strokeColor:'blue',straightFirst:false,straightLast:false,firstArrow:false,lastArrow:{type:2,size:4}}); ]]> 3.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:7}; n={-2,3}; bn={1:99999}; direction={0,1}; 0,"-","+"); disb=abs(b); chSign=["positive","negative"]; chDir=["up","down"]; ansB=pick(m>0,1,0);]]> none 0 1 0 3 [b,0,b] _err == 0 1 1 The order pair of the y-intercept is ({_1},{_2})
]]> The y-intercept is {b}.
The ordered pair of the y-intercept is (0,{b}).

]]> 1 1 0 2 [mNum,mDen] _err == 0 1 1 The slope is {_0} {_1} ]]> The slope is $\frac{{mNum}}{{mDen}}$.

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

]]> 2 1 0 4 [mNum,mDen,ansB,ansB] _err == 0 1 1 The change in y is {_0}.

The change in x is {_1}

The slope is {_2:chSign:MCE}, therefore go {_3:chDir:MCE} in the y-direction.

]]>

The change in y is {mNum}.

The change in x is {mDen}.

The slope is {=chSign[ansB]}, therefore go {=chDir[ansB]} in the y-direction.

]]> L84-Name slope Give the slope of the linear equation. y=$\frac{{mNum}}{{mDen}}$x + {b}

]]>

y=$\frac{{mNum}}{{mDen}}$x + {b}

The slope is $\frac{{mNum}}{{mDen}}$.

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:7}; bn={1:99999}; slope={-1,1}; 0,"less than zero, it is negative","greater than zero, it is positive"); m=mA[0]/mA[1]; MB=join("","MyBoxH1",bn); mNum=mA[0]; mDen=mA[1]; GCF=gcd(mNum,mDen); mNum=-1*mNum/GCF; mDen=mDen/GCF; m=mNum/mDen;]]> none 0 1 0 2 [mNum,mDen] _err == 0 1 1 The slope is {_0}{_1}]]> L84-Name slope (neg in front of slope) Give the slope of the linear equation. y={signm}$\frac{{DismNum}}{{mDen}}$x + {b}

]]>

y={signm} $\frac{{DismNum}}{{mDen}}$x + {b}

The slope is {signm} $\frac{{mNum}}{{mDen}}$.

{FB}

The change in y is {mNum}.	Slope =	{mNum}
The change in x is {mDen}.	Slope =	{mDen}

]]> 1.0000000 0.3333333 0 Your answer is correct. Your answer is partially correct. Your answer is incorrect. mA={[1,2],[1,3],[2,3],[1,4],[2,4],[3,4],[1,5],[2,5],[3,5],[1,6],[2,6],[3,6],[4,6],[5,6],[1,7],[2,7],[3,7],[4,7],[5,7],[6,7],[1,8],[2,8],[3,8],[4,8],[5,8],[6,8],[7,8],[1,10],[2,10],[3,10],[4,10],[5,10],[6,10],[7,10],[8,10],[9,10],[2,1],[3,1],[3,2],[4,1],[4,2],[4,3],[5,1],[5,2],[5,3],[6,1],[6,2],[6,3],[6,4]}; b={-7:7}; bn={1:99999}; slope={-1,1}; 0,"less than zero, it is negative","greater than zero, it is positive"); MB=join("","MyBoxH1",bn); mNum=mA[0]; mDen=mA[1]; GCF=gcd(mNum,mDen); mNum=slope*mNum/GCF; mDen=mDen/GCF; m=mNum/mDen; signm=pick(slope>=0,"-",""); DismNum=abs(mNum); FB=pick(slope>=0,"The slope is negative, place the negative sign in the numerator.","");]]> none 0 1 0 2 [mNum,mDen] _err == 0 1 1 The slope is {_0}{_1}]]> L84-Name y-Intercept Give the ordered pair of the y-intercept of the linear equation
y=$\frac{{mNum}}{{mDen}}$x {signB} {absB}

]]> The y-intercept is where the equation crosses the y-axis.
y={m}x {signB} {absB}
The y-intercept is at (0,{b}).

]]> The y-intercept is where the equation crosses the y-axis.
y={m}x {signB} {absB}
The identity property of addition is zero. adding zero does not change the sum.
The y-intercept is at (0,{b}).

TSA = 2\(\pi\)r² + \(\pi\)dh		Total Surface Area = The area of the two bases plus the area of the rectangle between them.
TSA = 2(3.14)({radius})² + (3.14)({diameter})({height})		Substitute the known values: Use \(\pi\) = 3.14 diameter = {diameter} Calculate the radius. Radius = \(\frac{1}{2}\) of the diameter = (\(\frac{1}{2}\))({diameter})={radius} height = {height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		{radius}² = {radius}{radius} = {=radiusradius} ({diameter})({height}) = {=diameter*height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		2(3.14)({=radiusradius}) = {=23.14radiusradius}
TSA = {=23.14radiusradius} + {=3.14diameter*height}		(3.14)({=diameterheight}) = {=3.14diameter*height}
TSA = {TSA} {unit}²		Add {=23.14radiusradius} and {=3.14diameterheight} {=23.14radiusradius} + {=3.14diameterheight} = {TSA}

TSA = 2\(\pi\)r² + \(\pi\)dh		Total Surface Area = The area of the two bases plus the area of the rectangle between them.
TSA = 2(3.14)({radius})² + (3.14)({diameter})({height})		Substitute the known values: Use \(\pi\) = 3.14 diameter = {diameter} Calculate the radius. Radius = \(\frac{1}{2}\) of the diameter = (\(\frac{1}{2}\))({diameter})={radius} height = {height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		{radius}² = {radius}{radius} = {=radiusradius} ({diameter})({height}) = {=diameter*height}
TSA = 2(3.14)({=radiusradius}) + (3.14)({=diameterheight})		2(3.14)({=radiusradius}) = {=23.14radiusradius}
TSA = {=23.14radiusradius} + {=3.14diameter*height}		(3.14)({=diameterheight}) = {=3.14diameter*height}
TSA = {TSA} {unit}²		Add {=23.14radiusradius} and {=3.14diameterheight} {=23.14radiusradius} + {=3.14diameterheight} = {TSA}

y=\(\frac{{mNum}}{{mDen}}\)x + {b}

y={signm} \(\frac{{DismNum}}{{mDen}}\)x + {b}

Area of the top and bottom bases
var board1=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-6,6,6,-6], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p4 = board1.create('point', [0,0],{fixed:true,size:'0',name:'',showInfobox:false}); var p5 = board1.create('point', [0, 5],{fixed:true,size:'0',name:'',showInfobox:false}); var base = board1.create('circle', [ p4, p5], {strokeColor:'green',fillColor:'pink'}); var tRadius=board1.create('text',[0,0.1,"{diameter} {unit}"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'20'}); var diameterL=board1.create('line',[[-5,0],[5,0]],{fixed:true,straightFirst:false,straightLast:false,dash:1,strokeColor:'red'});	Area = \(\pi\) radius²	The Formula for Area
	Area = (3.14)({radius})²	Substitute the known values: Use \(\pi\) = 3.14 diameter = {diameter}, therefore calculate the radius. Radius = \(\frac{1}{2}\) of the diameter = (\(\frac{1}{2}\))({diameter})={radius}
	Area = (3.14)({=radius*radius})	Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
	Area = {area} {unit}²	Multiply (3.14)({=radius*radius})
		*** Remember that Area is square units

Area of the rectangle (the section between the bases)
var board2=JXG.JSXGraph.initBoard(BOARDID,{ boundingbox:[-6,6,6,-6], axis:false, grid:false, showCopyright:false, showNavigation:false, keepaspectratio:true }); var p1 = board2.create('point', [-3,4],{fixed:true,size:'0',name:'',showInfobox:false}); var p2 = board2.create('point', [3, 4],{fixed:true,size:'0',name:'',showInfobox:false}); var p3 = board2.create('point', [3,-4],{fixed:true,size:'0',name:'',showInfobox:false}); var p4 = board2.create('point', [-3,-4],{fixed:true,size:'0',name:'', showInfobox:false}); var base = board2.create('polygon', [ p1, p2,p3,p4], {strokeColor:'black',fillColor:'blue'}); var tCircumference=board2.create('text',[0,4.1,"Circumference of the circle"],{fixed:true,anchorX:'middle',anchorY:'bottom',fontSize:'15'}); var tHeight=board2.create('text',[4.1,0,"{height} {unit}"],{fixed:true,anchorX:'left',anchorY:'middle',fontSize:'15'});	Area = lw Area = \(\pi\)dh	Formula for Area of a rectangle The length is the circumference of the circle (c=\(\pi\)d) The width is the height of the cylinder (h)
	Area = (3.14)({diameter})({height})	Substitute the known values: Use \(\pi\) = 3.14 diameter = {diameter} h is height = {height} {unit}
	Area= (3.14)({=diameter*height})	Calculate ({diameter})({height}) ({diameter})({height}) = {=diameter*height}
	*Area= {=cheight} {unit}²**	Multiply (3.14)({=diameter*height})

Volume of the Cylinder
V = \(\pi\)r²h		Formula for volume
Volume = (3.14)({radius})²{height}		Substitute the known values: Use \(\pi\) = 3.14 diameter = {diameter}, therefore calculate the radius. Radius = \(\frac{1}{2}\) of the diameter = (\(\frac{1}{2}\))({diameter})={radius} h is height = {height} {unit}
Volume = (3.14)({=radius*radius})({height})		Calculate ({radius})² ({radius})²=({radius})({radius}) = {=radius*radius}
*Volume = (3.14)({=radiusradiusheight})*		Multiply ({=radius*radius}) times {height}
Volume = {volume} {unit}³		Multiply (3.14)({=radiusradiusheight})