Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
Add {=floor(m)} + {=round((missing-1)*m,0)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {missing} = {=floor(m) + round((missing-1)*m,0)}
A reasonable answer is {=round((missing*m)-m*.5,0)} to {=round((missing*m)+m*.5,0)}
]]>Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
{=XTextUnit[randX]} 1: {=yTextAlc[randY]} = {=floor(m)}
{=XTextUnit[randX]} 2: {=yTextAlc[randY]} = {=floor(2*m)}
{=XTextUnit[randX]} {=maxX}: {=yTextAlc[randY]} = {=floor((maxX-1)*m)}
Add {=floor(m)} + {=floor((maxX-1)*m)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {=maxX+1} = {=round(maxValue+m,0)}
A reasonable answer is {=round(maxValue+(.5*m),0)} - {=round(maxValue+(1.5*m),0)}
]]>student # {_0} received a {=factor*score}%
]]>Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
Add {=floor(m)} + {=round((missing-1)*m,0)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {missing} = {=floor(m) + round((missing-1)*m,0)}
A reasonable answer is {=round((missing*m)-m*.5,0)} to {=round((missing*m)+m*.5,0)}
]]>Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
{=XTextUnit[randX]} 1: {=yTextAlc[randY]} = {=floor(m)}
{=XTextUnit[randX]} 2: {=yTextAlc[randY]} = {=floor(2*m)}
{=XTextUnit[randX]} {=maxX}: {=yTextAlc[randY]} = {=floor((maxX-1)*m)}
Add {=floor(m)} + {=floor((maxX-1)*m)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {=maxX+1} = {=round(maxValue+m,0)}
A reasonable answer is {=round(maxValue+(.5*m),0)} - {=round(maxValue+(1.5*m),0)}
]]>student # {_0} received a {=factor*score}%
]]>The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The students favorite activity is represented by the {=randColors[randHigh]} section.
The {=randColors[randHigh]} section of the pie chart is the largest section with {DataH} votes.
The activity is {=randE[randHigh]}.
The students least favorite activity is represented by the {=randColors[randLow]} section.
The {=randColors[randLow]} section of the pie chart is the smallest section section with {DataL} votes.
The activity is {=randE[randLow]}.
The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The students favorite {activity} is represented by the {=randColors[randHigh]} section.
The {=randColors[randHigh]} section of the pie chart is the largest section with {DataH} votes.
The {activity} is {=randE[randHigh]}.
The students least favorite {activity} is represented by the {=randColors[randLow]} section.
The {=randColors[randLow]} section of the pie chart is the smallest section section with {DataL} votes.
The {activity} is {=randE[randLow]}.
Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
Add {=floor(m)} + {=round((missing-1)*m,0)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {missing} = {=floor(m) + round((missing-1)*m,0)}
A reasonable answer is {=round((missing*m)-m*.5,0)} to {=round((missing*m)+m*.5,0)}
]]>Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
{=XTextUnit[randX]} 1: {=yTextAlc[randY]} = {=floor(m)}
{=XTextUnit[randX]} 2: {=yTextAlc[randY]} = {=floor(2*m)}
{=XTextUnit[randX]} {=maxX}: {=yTextAlc[randY]} = {=floor((maxX-1)*m)}
Add {=floor(m)} + {=floor((maxX-1)*m)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {=maxX+1} = {=round(maxValue+m,0)}
A reasonable answer is {=round(maxValue+(.5*m),0)} - {=round(maxValue+(1.5*m),0)}
]]>student # {_0} received a {=factor*score}%
]]>The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The students favorite activity is represented by the {=randColors[randHigh]} section.
The {=randColors[randHigh]} section of the pie chart is the largest section with {DataH} votes.
The activity is {=randE[randHigh]}.
The students least favorite activity is represented by the {=randColors[randLow]} section.
The {=randColors[randLow]} section of the pie chart is the smallest section section with {DataL} votes.
The activity is {=randE[randLow]}.
The {=randColors[randHigh-1]} section of the pie chart is the largest section.
]]>The students favorite {activity} is represented by the {=randColors[randHigh]} section.
The {=randColors[randHigh]} section of the pie chart is the largest section with {DataH} votes.
The {activity} is {=randE[randHigh]}.
The students least favorite {activity} is represented by the {=randColors[randLow]} section.
The {=randColors[randLow]} section of the pie chart is the smallest section section with {DataL} votes.
The {activity} is {=randE[randLow]}.
Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
Add {=floor(m)} + {=round((missing-1)*m,0)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {missing} = {=floor(m) + round((missing-1)*m,0)}
A reasonable answer is {=round((missing*m)-m*.5,0)} to {=round((missing*m)+m*.5,0)}
]]>Each {=XTextUnit[randX]} the {=yTextAlc[randY]} increases by approximately {=floor(m)}
{=XTextUnit[randX]} 1: {=yTextAlc[randY]} = {=floor(m)}
{=XTextUnit[randX]} 2: {=yTextAlc[randY]} = {=floor(2*m)}
{=XTextUnit[randX]} {=maxX}: {=yTextAlc[randY]} = {=floor((maxX-1)*m)}
Add {=floor(m)} + {=floor((maxX-1)*m)}
The {=yTextAlc[randY]} in {=XTextUnit[randX]} {=maxX+1} = {=round(maxValue+m,0)}
A reasonable answer is {=round(maxValue+(.5*m),0)} - {=round(maxValue+(1.5*m),0)}
]]>student # {_0} received a {=factor*score}%
]]>
Area = ({length})({width})
Area = {A} {=unit[1]}
]]>The length was {length}.
{=factorW[factor]} means {factor} times.
{length} x {factor} = {=length*factor}
]]>Area of the larger pen = length x width
Area of the larger pen= ({=length*factor})({width})
Area of the larger pen= {An} {=unit[1]}
Area = length x width
A={l} x {w} = {areaP}
Double the length: 2 x {l} = {=2*l}
Double the width: 2 x {w} = {=2*w}
Area of the larger pen = {=2*l} * {=2*w} = {areaLP}
{num1} {=operations[op0]} {num2} {=operations[op1]} {num3} {=operations[op2]} {num4} {=operations[op3]} {num5}={ans}
{=opWord[op0]}: {num1} {=operations[op0]} {num2} = {s1}
{num1} {=operations[op0]} ( {num2} {=operations[op1]} {num3}) {=operations[op2]} {num4}={ans}
Simplify inside parentheses first: {num2} {=operations[op1]} {num3} = {s1}
|
Write the answer in simplest form. DO NOT put a zero in the denominator. Use the denominator from the question) |
{#A} |
The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
|
\(\frac{{factor0}}{{factor0}}\) = 1 \( \frac{{num0}}{{den0}} \) • \(\frac{{factor0}}{{factor0}}\) = \( \frac{{ansNum0}}{{ansDen0}} \) |
\(\frac{{factor1}}{{factor1}}\) = 1 \( \frac{{num1}}{{den1}} \) • \(\frac{{factor1}}{{factor1}}\) = \( \frac{{ansNum1}}{{ansDen1}} \) |
Add the numerators: {ansNum0}+{ansNum1} = {impN}
Keep the denominator: {least}
The sum of the fractions is \(\frac{{impN}}{{least}}\)
{FB1}
The fractional part is \(\frac{{ansN}}{{ansD}}\)
Determine the GCF of the numerator and denominator:
The factors of {ansN}: {=FB[ansN]}
The factors of {ansD}: {=FB[ansD]}
The GCF (greatest common factor) of {ansN} and {ansD} is {gcfND}.
| {ansN} | ÷ |
{gcfND} | = | {ansN}÷{gcfND} | = | {=ansN/gcfND} |
| {ansD} | {gcfND} | {ansD}÷{gcfND} | {=ansD/gcfND} |
The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
]]>|
Write the answer in simplest form. DO NOT put a zero in the denominator. Use the denominator from the question) |
{#A} |
The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
|
\(\frac{{factor0}}{{factor0}}\) = 1 \( \frac{{num0}}{{den0}} \) • \(\frac{{factor0}}{{factor0}}\) = \( \frac{{ansNum0}}{{ansDen0}} \) |
\(\frac{{factor1}}{{factor1}}\) = 1 \( \frac{{num1}}{{den1}} \) • \(\frac{{factor1}}{{factor1}}\) = \( \frac{{ansNum1}}{{ansDen1}} \) |
Add the numerators: {ansNum0}+{ansNum1} = {impN}
Keep the denominator: {least}
The sum of the fractions is \(\frac{{impN}}{{least}}\)
{FB1}
The fractional part is \(\frac{{ansN}}{{ansD}}\)
Determine the GCF of the numerator and denominator:
The factors of {ansN}: {=FB[ansN]}
The factors of {ansD}: {=FB[ansD]}
The GCF (greatest common factor) of {ansN} and {ansD} is {gcfND}.
| {ansN} | ÷ |
{gcfND} | = | {ansN}÷{gcfND} | = | {=ansN/gcfND} |
| {ansD} | {gcfND} | {ansD}÷{gcfND} | {=ansD/gcfND} |
The decimal part is read {=wordA[tenths]}-{=word[hundredths]} hundredths.
Write the decimal part as a fraction \(\frac{{decNum}}{100}\).
Simplify the fraction by dividing the numerator and denominator by the GCF:
\(\frac{{tenths}{hundredths}÷{GCF}}{{place}÷{GCF}}\) = \(\frac{{num}}{{den}}\)
The answer is {whole}.{tenths}{hundredths} = \(\frac{{num}}{{den}}\)
]]>The decimal part is read {=word[tenths]} tenths.
Write the decimal part as a fraction \(\frac{{tenths}}{10}\).
Simplify the fraction by dividing the numerator and denominator by the GCF:
\(\frac{{tenths}÷{GCF}}{10÷{GCF}}\) = \(\frac{{num}}{{den}}\)
The answer is {whole}.{tenths} = \(\frac{{num}}{{den}}\)
]]>The decimal part is read {=wordA[tenths]}-{=word[hundredths]} hundredths.
Write the decimal part as a fraction \(\frac{{decNum}}{100}\).
Simplify the fraction by dividing the numerator and denominator by the GCF:
\(\frac{{tenths}{hundredths}÷{GCF}}{{place}÷{GCF}}\) = \(\frac{{num}}{{den}}\)
The answer is {whole}.{tenths}{hundredths} = \(\frac{{num}}{{den}}\)
]]>The decimal part is read {=wordA[tenths]} hundred {=word[hundredths]}-{=word[thousandths]} thousandths.
Write the decimal part as a fraction \(\frac{{decNum}}{1000}\).
Simplify the fraction by dividing the numerator and denominator by the GCF:
\(\frac{{tenths}{hundredths}{thousandths}÷{GCF}}{{place}÷{GCF}}\) = \(\frac{{num}}{{den}}\)
The answer is {whole}.{tenths}{hundredths}{thousandths} = \(\frac{{num}}{{den}}\)
]]>| {whole}.{tenths}{hundredths}{thousandths} ={_0} | {_1} |
| {_2} |
| {whole}.{tenths}{hundredths}{thousandths} ={_0} | {_1} |
| {_2} |
The decimal part is read {=word[tenths]} tenths.
Write the decimal part as a fraction \(\frac{{tenths}}{10}\).
Simplify the fraction by dividing the numerator and denominator by the GCF:
\(\frac{{tenths}÷{GCF}}{10÷{GCF}}\) = \(\frac{{num}}{{den}}\)
The answer is {whole}.{tenths} = {whole}\(\frac{{num}}{{den}}\)
]]>| {p1} | Given | |
| {p1S1} | {p1S1FB} | |
| {p1S2} | {p1S2FB} |
| {p2} | Given | |
| {p2S1} | {p2S1FB} | |
| {p2S2} | {p2S2FB} |
| {p3} | Given | |
| {p3S1} | {p3S1FB} | |
| {p3S2} | {p3S2FB} |
| {p4} | Given | |
| {p4S1} | {p4S1FB} | |
| {p4S2} | {p4S2FB} |
The first point can be at x= {x2}.
When x= {x2} y={y2}
The first point can be ({x2},{y2})
The second point can be at x= {x3}
When x= {x3} y={y3}
The second point can be ({x3},{y3})
Use one of the other two point to check the answer.
| {p1} | Given | |
| {p1S1} | {p1S1FB} | |
| {p1S2} | {p1S2FB} |
| {p2} | Given | |
| {p2S1} | {p2S1FB} | |
| {p2S2} | {p2S2FB} |
| {p3} | Given | |
| {p3S1} | {p3S1FB} | |
| {p3S2} | {p3S2FB} |
| {p4} | Given | |
| {p4S1} | {p4S1FB} | |
| {p4S2} | {p4S2FB} |
The first point can be at x= {x2}.
When x= {x2} y={y2}
The first point can be ({x2},{y2})
The second point can be at x= {x3}
When x= {x3} y={y3}
The second point can be ({x3},{y3})
Use one of the other two point to check the answer.
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({base})({width}) | Use parentheses to show multiplication | |
| A = {area} square feet | Use square units for area. You are finding the number of square units inside of the polygon. | |
| The perimeter of a parallelogram is the sum of the distance of each side | ||
| P = side + side + base + base P=s+s+b+b P = 2s + 2b |
Formulas are usually written with the abbreviations A is area = s means side, b means base and h means height side + side is the same as 2s (2 times the side) and base + base is the same as 2b (2 times the base) |
|
| P = 2({side})+2({base}) | Use parentheses to show multiplication | |
| P = {=2*side}+{=2*base} | 2({side}) = {=2*side} and 2({base}) = {=2*base} | |
| P = {perimeter} {unit} | Perimeter is the distance around a figure. Since a ruler is used to measure distance, use {unit}. | |
| The perimeter of a parallelogram is the sum of the distance of each side | ||
| P = side + side + base + base P=s+s+b+b P = 2s + 2b |
Formulas are usually written with the abbreviations A is area = s means side, b means base and h means height side + side is the same as 2s (2 times the side) and base + base is the same as 2b (2 times the base) |
|
| P = 2({side})+2({base}) | Use parentheses to show multiplication | |
| P = {=2*side}+{=2*base} | 2({side}) = {=2*side} and 2({base}) = {=2*base} | |
| P = {perimeter} {unit} | Perimeter is the distance around a figure. Since a ruler is used to measure distance, use {unit}. | |
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({base})({width}) | Use parentheses to show multiplication | |
| A = {area} square feet | Use square units for area. You are finding the number of square units inside of the polygon. | |
A=\(\frac{1}{2}\)({length} x {width})
A=\(\frac{1}{2}\)(\(\frac{{=length*width}}{1}\))
A={=length*width/2} {=units[1]}
Area of a Triangle = base x height ÷ 2
A = {length} x {width} ÷ 2
A = {=length*width} ÷ 2
A = {area} {=units[1]}
The division can be written with the fraction bar.
A = \(\frac{({length})({width})}{2}\)
A = \(\frac{({=length*width})}{2}\)
A={area} {=units[1]}
\(\frac{{num1}}{{den}}\) of {number}
\(\frac{{num1}}{{den}}\) x {number}
\(\frac{{num1}}{{den}}\) of {number} = {ans1}
\(\frac{{num2}}{{den}}\) of {number} | Given | |
\(\frac{{num2}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num2}}{{den}}\) x {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num2} groups. {num2} x {ans1} = {ans2} The answer is {ans2}. | |
\(\frac{{num3}}{{den}}\) of {number} | Given | |
\(\frac{{num3}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num3}}{{den}}\) of {number} = {ans3} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num3} groups. {num3} x {ans1} = {ans3} The answer is {ans3}. | |
\(\frac{{num4}}{{den}}\) of {number} | Given | |
\(\frac{{num4}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num4}}{{den}}\) of {number} = {ans4} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num4} groups. {num4} x {ans1} = {ans4} The answer is {ans4}. | |
\(\frac{{num5}}{{den}}\) of {number} | Given | |
\(\frac{{num5}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num5}}{{den}}\) of {number} = {ans5} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num5} groups. {num5} x {ans1} = {ans5} The answer is {ans5}. | |
\(\frac{{num1}}{{den}}\) of {number}
\(\frac{{num1}}{{den}}\) x {number}
\(\frac{{num1}}{{den}}\) of {number} = {ans1}
\(\frac{{num2}}{{den}}\) of {number} | Given | |
\(\frac{{num2}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num2}}{{den}}\) x {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num2} groups. {num2} x {ans1} = {ans2} The answer is {ans2}. | |
\(\frac{{num3}}{{den}}\) of {number} | Given | |
\(\frac{{num3}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num3}}{{den}}\) of {number} = {ans3} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num3} groups. {num3} x {ans1} = {ans3} The answer is {ans3}. | |
\(\frac{{num4}}{{den}}\) of {number} | Given | |
\(\frac{{num4}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num4}}{{den}}\) of {number} = {ans4} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num4} groups. {num4} x {ans1} = {ans4} The answer is {ans4}. | |
\(\frac{{num1}}{{den}}\) of {number}
\(\frac{{num1}}{{den}}\) x {number}
\(\frac{{num1}}{{den}}\) of {number} = {ans1}
\(\frac{{num2}}{{den}}\) of {number} | Given | |
\(\frac{{num2}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num2}}{{den}}\) x {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num2} groups. {num2} x {ans1} = {ans2} The answer is {ans2}. | |
\(\frac{{num3}}{{den}}\) of {number} | Given | |
\(\frac{{num3}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num3}}{{den}}\) of {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num3} groups. {num3} x {ans1} = {ans3} The answer is {ans3}. | |
Area of a Triangle = base x height ÷ 2
A = {length} x {width} ÷ 2
A = {=length*width} ÷ 2
A = {area} {=units[1]}
The division can be written with the fraction bar.
A = \(\frac{({length})({width})}{2}\)
A = \(\frac{({=length*width})}{2}\)
A={area} {=units[1]}
(Leave the answer as an improper fraction.)
]]>| {num1} | x | {impNum2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
(Leave the answer as an improper fraction.)
]]>| {impNum1} | x | {num2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
(Leave the answer as an improper fraction.)
]]>| {impNum1} | x | {impNum2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
(\(\frac{{num}}{{den}}\))({whole1}\(\frac{{num1}}{{den1}}\))=
(\(\frac{{num}}{{den}}\))({whole}) =
| {num1} | x | {num2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
({whole}\(\frac{{num}}{{den}}\))(\(\frac{{num1}}{{den1}}\))=
({whole}\(\frac{{num}}{{den}}\))({whole1}\(\frac{{num1}}{{den1}}\))=
(Leave the answer as an improper fraction.)
]]>| {impNum1} | x | {num2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
({whole})(\(\frac{{num}}{{den}}\))=
{first} {units[0]} : {second} {units[1]}
\(\frac{{first}}{{GCF}}\) : \(\frac{{second}}{{GCF}}\)
{a} {units[0]} : {den} {units[1]}
| {first} {units[0]} |
| {second} {units[1]} |
| {first} ÷ {GCF} |
| {second} ÷ {GCF} |
| {a} {units[0]} |
| {den} {units[1]} |
{first} {units[0]} : {second} {units[1]}
{units[2]} = {=first+second}
{first} {units[0]} : {=first+second} {units[2]}
\(\frac{{first}}{{GCFb}}\) : \(\frac{{=first+second}}{{GCFb}}\)
{b} {units[0]} : {totb} {units[2]}
{first} {units[0]} : {second} {units[1]}
{units[2]} = {=first+second}
| {first} {units[0]} |
| {=first+second} {units[2]} |
| {first}÷{GCFb} |
| {=first+second}÷{GCFb} |
| {b} {units[0]} |
| {totb} {units[2]} |
{first} {units[0]} to {second} {units[1]}
{units[2]} = {=first+second}
{first} {units[0]} to {=first+second} {units[2]}
\(\frac{{first}}{{GCFb}}\) to \(\frac{{=first+second}}{{GCFb}}\)
{b} {units[0]} to {totb} {units[2]}
The order of the relationship is important. Write {=scenario[0]} on the top and {=scenario[1]} on the bottom, or under the fraction bar.
\(\frac{{ratio0} {=scenario[0]}}{{ratio1} {=scenario[1]}}\)
{FB}
\(\frac{{ratio0} ÷ {GCF}}{{ratio1} ÷ {GCF}}\)
\(\frac{{num1}}{{num2}}\)
The order of the relationship is important. Write {=scenario[0]} on the left and {=scenario[1]} on the right of the colon.
{ratio0} {=scenario[0]} : {ratio1} {=scenario[1]}
{FB}
{ratio0} ÷ {GCF} : {ratio1} ÷ {GCF}
{num1} : {num2}
The order of the relationship is important. Write {=scenario[0]} on the left and {=scenario[1]} on the right of the word "to".
{ratio0} {=scenario[0]} to {ratio1} {=scenario[1]}
{FB}
{ratio0} ÷ {GCF} to {ratio1} ÷ {GCF}
{num1} to {num2}
{first} {units[0]} {units[1]} {second} {units[2]}
\(\frac{{first}}{{GCF}}\) : \(\frac{{second}}{{GCF}}\)
{a} {units[0]} : {den} {units[2]}
| {first} {units[0]} |
| {second} {units[2]} |
| {first} ÷ {GCF} |
| {second} ÷ {GCF} |
| {a} {units[0]} |
| {den} {units[2]} |
| Ω = {a} | Ω - θ = {=a-b} | |
| θ + α = {=d} | α + α = {=c+c} |
| Ω = {a} | Ω - θ = {=a-b} | |
| θ + α = {d} | α + α = {=c+c} |
Ω = {_0}
θ = {_1}
α = {_2}
| {bfac}Ω = θ | Ω + θ = {=a+b} | |
| {k1} + α = {=k1+c} | Ω + α = {=a+c} |
| {bfac}Ω = θ | Ω + θ = {=a+b} | |
| {k1} + α = {=k1+c} | Ω + α = {=a+c} |
Ω = {_0}
θ = {_1}
α = {_2}
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({base})({width}) | Use parentheses to show multiplication | |
| A = {area} square {unit} | Use square units for area. You are finding the number of square units inside of the polygon. | |
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({base})({width}) | Use parentheses to show multiplication | |
| A = {area} square feet | Use square units for area. You are finding the number of square units inside of the polygon. | |
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({length})({width}) | Use parentheses to show multiplication | |
| A = {area} square {unit} | Use square units for area. You are finding the number of square units inside of the polygon. | |
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({base})({width}) | Use parentheses to show multiplication | |
| A = {area} square feet | Use square units for area. You are finding the number of square units inside of the polygon. | |
| The area or a parallelogram= base times height | ||
| A = bh | Formulas are usually written with the abbreviations A is area = b means base and h means height | |
| A = ({length})({width}) | Use parentheses to show multiplication | |
| A = {area} square feet | Use square units for area. You are finding the number of square units inside of the polygon. | |
The side of the parallelogram = {side} {unit}, the base = {base} {unit}, and the height = {width} {unit}
| The perimeter of a parallelogram is the sum of the distance of each side | ||
| P = side + side + base + base P=s+s+b+b P = 2s + 2b |
Formulas are usually written with the abbreviations A is area = s means side, b means base and h means height side + side is the same as 2s (2 times the side) and base + base is the same as 2b (2 times the base) |
|
| P = 2({side})+2({base}) | Use parentheses to show multiplication | |
| P = {=2*side}+{=2*base} | 2({side}) = {=2*side} and 2({base}) = {=2*base} | |
| P = {perimeter} {unit} | Perimeter is the distance around a figure. Since a ruler is used to measure distance, use {unit}. | |
The side of the parallelogram = {side} {unit}, the base = {base} {unit}, and the height = {width} {unit}
| The perimeter of a parallelogram is the sum of the distance of each side | ||
| P = side + side + base + base P=s+s+b+b P = 2s + 2b |
Formulas are usually written with the abbreviations A is area = s means side, b means base and h means height side + side is the same as 2s (2 times the side) and base + base is the same as 2b (2 times the base) |
|
| P = 2({side})+2({base}) | Use parentheses to show multiplication | |
| P = {=2*side}+{=2*base} | 2({side}) = {=2*side} and 2({base}) = {=2*base} | |
| P = {perimeter} {unit} | Perimeter is the distance around a figure. Since a ruler is used to measure distance, use {unit}. | |
(Leave the answer as an improper fraction.)
]]>| {num1} | x | {impNum2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
(Leave the answer as an improper fraction.)
]]>| {impNum1} | x | {num2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
(Leave the answer as an improper fraction.)
]]>| {impNum1} | x | {impNum2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
(\(\frac{{num}}{{den}}\))({whole1}\(\frac{{num1}}{{den1}}\))=
({whole}\(\frac{{num}}{{den}}\))(\(\frac{{num1}}{{den1}}\))=
({whole}\(\frac{{num}}{{den}}\))({whole1}\(\frac{{num1}}{{den1}}\))=
(Leave the answer as an improper fraction.)
]]>| {impNum1} | x | {num2} | = | {prodNum} |
| {den1} | {den2} | {prodDen} |
A=\(\frac{1}{2}\)({length} x {width})
A=\(\frac{1}{2}\)(\(\frac{{=length*width}}{1}\))
A={=length*width/2} {=units[1]}
Area of a Triangle = base x height ÷ 2
A = {length} x {width} ÷ 2
A = {=length*width} ÷ 2
A = {area} {=units[1]}
The division can be written with the fraction bar.
A = \(\frac{({length})({width})}{2}\)
A = \(\frac{({=length*width})}{2}\)
A={area} {=units[1]}
Area of a Triangle = base x height ÷ 2
A = {length} x {width} ÷ 2
A = {=length*width} ÷ 2
A = {area} {=units[1]}
The division can be written with the fraction bar.
A = \(\frac{({length})({width})}{2}\)
A = \(\frac{({=length*width})}{2}\)
A={area} {=units[1]}
(\(\frac{{num}}{{den}}\))({whole}) =
({whole})(\(\frac{{num}}{{den}}\))=
\(\frac{{num1}}{{den}}\) of {number}
\(\frac{{num1}}{{den}}\) x {number}
\(\frac{{num1}}{{den}}\) of {number} = {ans1}
\(\frac{{num2}}{{den}}\) of {number} | Given | |
\(\frac{{num2}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num2}}{{den}}\) x {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num2} groups. {num2} x {ans1} = {ans2} The answer is {ans2}. | |
\(\frac{{num3}}{{den}}\) of {number} | Given | |
\(\frac{{num3}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num3}}{{den}}\) of {number} = {ans3} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num3} groups. {num3} x {ans1} = {ans3} The answer is {ans3}. | |
\(\frac{{num4}}{{den}}\) of {number} | Given | |
\(\frac{{num4}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num4}}{{den}}\) of {number} = {ans4} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num4} groups. {num4} x {ans1} = {ans4} The answer is {ans4}. | |
\(\frac{{num5}}{{den}}\) of {number} | Given | |
\(\frac{{num5}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num5}}{{den}}\) of {number} = {ans5} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num5} groups. {num5} x {ans1} = {ans5} The answer is {ans5}. | |
\(\frac{{num1}}{{den}}\) of {number}
\(\frac{{num1}}{{den}}\) x {number}
\(\frac{{num1}}{{den}}\) of {number} = {ans1}
\(\frac{{num2}}{{den}}\) of {number} | Given | |
\(\frac{{num2}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num2}}{{den}}\) x {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num2} groups. {num2} x {ans1} = {ans2} The answer is {ans2}. | |
\(\frac{{num3}}{{den}}\) of {number} | Given | |
\(\frac{{num3}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num3}}{{den}}\) of {number} = {ans3} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num3} groups. {num3} x {ans1} = {ans3} The answer is {ans3}. | |
\(\frac{{num4}}{{den}}\) of {number} | Given | |
\(\frac{{num4}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num4}}{{den}}\) of {number} = {ans4} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num4} groups. {num4} x {ans1} = {ans4} The answer is {ans4}. | |
\(\frac{{num1}}{{den}}\) of {number}
\(\frac{{num1}}{{den}}\) x {number}
\(\frac{{num1}}{{den}}\) of {number} = {ans1}
\(\frac{{num2}}{{den}}\) of {number} | Given | |
\(\frac{{num2}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num2}}{{den}}\) x {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num2} groups. {num2} x {ans1} = {ans2} The answer is {ans2}. | |
\(\frac{{num3}}{{den}}\) of {number} | Given | |
\(\frac{{num3}}{{den}}\) x {number} | of means multiply | |
\(\frac{{num3}}{{den}}\) of {number} = {ans2} | Separate {number} into {den} equal groups. Each group has {ans1} items. The numerator tells us to take {num3} groups. {num3} x {ans1} = {ans3} The answer is {ans3}. | |