]]>The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
]]>The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
]]>| {number1} | , | {#num12} | , | {#num13} | , | {#num14} | , | {#num15} | , | {#num16} | , | {#num17} | , | {#num18} | , | {#num19} | , | {#num110} |
What are the multiples of {number2}?
| {number2} | , | {#num22} | , | {#num23} | , | {#num24} | , | {#num25} | , | {#num26} | , | {#num27} | , | {#num28} | , | {#num29} | , | {#num210} |
What is the least common multiple of {number1} and {number2}?
{#least}
]]>The least common multiple of {number1} and {number2} is {least}
The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
The multiples of {number1} are: {multiples1}
The multiples of {number2} are: {multiples2}
]]>]]>Add the frequency of each score:
| Score | Frequency | |
| {values[0]} | {freq[0]} | |
| {values[1]} | {freq[1]} | |
| {values[2]} | {freq[2]} | |
| {values[3]} | {freq[3]} | |
| {values[4]} | {freq[4]} | |
| {values[5]} | {freq[5]} | |
| {values[6]} | {freq[6]} | |
| {values[7]} | {freq[7]} | |
| {values[8]} | {freq[8]} | |
| {values[9]} | {freq[9]} | |
| {values[10]} | {freq[10]} |
Add the frequency of each score:
| Score | Frequency | |
| {values[0]} | {freq[0]} | |
| {values[1]} | {freq[1]} | |
| {values[2]} | {freq[2]} | |
| {values[3]} | {freq[3]} | |
| {values[4]} | {freq[4]} | |
| {values[5]} | {freq[5]} | |
| {values[6]} | {freq[6]} | |
| {values[7]} | {freq[7]} | |
| {values[8]} | {freq[8]} | |
| {values[9]} | {freq[9]} | |
| {values[10]} | {freq[10]} |
Add the frequency of each {record}:
| {record} | Frequency | |
| {values[0]} | {freq[0]} | |
| {values[1]} | {freq[1]} | |
| {values[2]} | {freq[2]} | |
| {values[3]} | {freq[3]} | |
| {values[4]} | {freq[4]} | |
| {values[5]} | {freq[5]} | |
| {values[6]} | {freq[6]} | |
| {values[7]} | {freq[7]} | |
| {values[8]} | {freq[8]} | |
| {values[9]} | {freq[9]} | |
| {values[10]} | {freq[10]} |
Add the frequency of each dice roll.
| dice roll | Frequency | |
| {values[0]} | {freq[0]} | |
| {values[1]} | {freq[1]} | |
| {values[2]} | {freq[2]} | |
| {values[3]} | {freq[3]} | |
| {values[4]} | {freq[4]} | |
| {values[5]} | {freq[5]} | |
| {values[6]} | {freq[6]} | |
| {values[7]} | {freq[7]} | |
| {values[8]} | {freq[8]} | |
| {values[9]} | {freq[9]} | |
| {values[10]} | {freq[10]} |
The fraction in simplest form is: \(\frac{{redNum}}{{redDen}}\)
(If the answer is a whole number use a "0" in the numerator)
{impNum} ÷ {denominator} = {wholeNum}
How many are left over? {mixNum} left over
| {wholeNum} | remainder {mixNum} | |
| {denominator} | {impNum} |
| {wholeNum} | + {redNum} | = | {impNum} |
| • {redDen} | {impDen} |
Multiply the denominator by the whole number: {redDen} • {wholeNum} = {=wholeNum*redDen}
Add the numerator to the product: {=wholeNum*redDen} + {redNum} = {impNum}
Keep the same denominator: {impDen}
The fraction in simplest form is: \(\frac{{redNum}}{{redDen}}\)
The fraction in simplest form is: \(\frac{{redNum}}{{redDen}}\)
The fraction is to the right of {wholeNum}, therefore the point is at {wholeNum}\(\frac{{redNum}}{{redDen}}\)
Round to the nearest whole number:
\(\frac{{redNum}}{{redDen}}\) {FBfrac}
What improper fraction represents the amount of circle shaded? |
{_0} |
{_1} |
The circle is divided into eight equal sections. Each section represents \(\frac{1}{8}\).
There are {num} sections shaded in the left circle.
{FBWhole}. Each whole circle represents \(\frac{8}{8}\).
Altogether there are {ansNum} eighths shaded.
The fraction is \(\frac{{ansNum}}{{den}}\).
Write the answer in lowest terms.
What mixed number represents the part of the shaded circle? |
{_0} |
{_1} |
{_2} |
Write the fraction in lowest terms.
The GCF of {num} and {den} = {gcfFrac}
A fraction can be reduced if the GCF is greater than 1.
To reduce the fraction divide the numerator and denominator each by the GCF
{num}/{gcfFrac} = {redMNum}
{den}/{gcfFrac}={redMDen}
The answer as a Mixed Number in simplest form is: {whole}\(\frac{{redMNum}}{{redMDen}}\)
What improper fraction represents the amount of circle shaded? |
{_0} |
{_1} |
The circle is divided into eight equal sections. Each section represents \(\frac{1}{{den}}\).
There are {num} sections shaded in the left circle.
{FBWhole}. Each whole circle represents \(\frac{{den}}{{den}}\).
Altogether there are {ansNum} {denWord} shaded.
The fraction is \(\frac{{ansNum}}{{den}}\).
Write the answer in lowest terms.
What mixed number represents the part of the shaded circle? |
{_0} |
{_1} |
{_2} |
Write the fraction in lowest terms.
The GCF of {num} and {den} = {gcfFrac}
A fraction can be reduced if the GCF is greater than 1.
To reduce the fraction divide the numerator and denominator each by the GCF
{num}/{gcfFrac} = {redMNum}
{den}/{gcfFrac}={redMDen}
The answer as a Mixed Number in simplest form is: {whole}\(\frac{{redMNum}}{{redMDen}}\)
What improper fraction represents the amount of circle shaded? |
{_0} |
{_1} |
The circle is divided into {numberWord} equal sections. Each section represents \(\frac{1}{{den}}\).
There are {num} sections shaded in the left circle.
{FBWhole}. Each whole circle represents \(\frac{{den}}{{den}}\).
Altogether there are {ansNum} {denWord} shaded.
The fraction is \(\frac{{ansNum}}{{den}}\).
Write the answer in lowest terms.
What mixed number represents the part of the shaded circle? |
{_0} |
{_1} |
{_2} |
Write the fraction in lowest terms.
The GCF of {num} and {den} = {gcfFrac}
A fraction can be reduced if the GCF is greater than 1.
To reduce the fraction divide the numerator and denominator each by the GCF
{num}/{gcfFrac} = {redMNum}
{den}/{gcfFrac}={redMDen}
The answer as a Mixed Number in simplest form is: {whole}\(\frac{{redMNum}}{{redMDen}}\)
| {#a} |
| {#a} |
| {#b} |
| {#c} |
| {#d} |
| {#a} |
| {#b} |
| {#c} |
| {#d} |
({cm})(10) = {mm} mm
]]>({m})(100) = {cm} cm
]]>{mm}÷10 = {cm} cm
]]>{startV} {startU} {FBsymbol} {factor} = {endV} {endU}
[[1]], [[2]], decimeter, [[3]], dekameter, hectometer, [[4]]
]]><A: {=choice[angA]} angle.
]]><B = {b}
<B: {=choice[angB]} angle.
<C = {c}
<C: {=choice[angC]} angle.
| x + {s} + 90 = \({180}^\circ\) | The sum of the angles = \({180}^\circ\) | |
| x + {=s+90} = \({180}^\circ\) | Simplify on the left side of the equation | |
| - {=s+90} = - {=s+90} |
Subtract {=s+90} from both sides | |
| x = \({{t}}^\circ\) | \(\angle\) A = \({{t}}^\circ\) |
| x + {t} + 90 = \({180}^\circ\) | The sum of the angles = \({180}^\circ\) | |
| x + {=90+t} = 180 | Add {t} + 90 | |
| -{=90+t} -{=90+t} | Subtract -{=90+t} from both sides | |
| x = {s} | Simplify | |
| \(\angle\)A = {s}°. |
Classify the triangle by its sides.
]]>< ABC is a(n) {_0:angle:MCE}, it measures {_1:degAns:MCE} degrees.
]]>]]>The factor must be between 1 and 10: {number}
The power of 10: 10{exponent} because the decimal moves {exponent} places.
]]>The factor must be between 1 and 10: {number}
The power of 10: 10{exponent} because the decimal moves {exponent} places.
]]>The factor is between 1 and 10.
Move the decimal {exponent} places to the right since 10{exponent} is {=pow(10,exponent)}.
]]>| {#a} |
| {#b} |
| {#c} |
| {#d} |
| {#a} |
| {#b} |
| {#c} |
| {#d} |
| {#a} |
| {#a} |
mm |
cm |
m |
km |
{cm}x10 {mm} |
{m}x100 {cm} |
{km}x1000 {m} |
{km} |
| mm | cm | m | km |
| {_0} | {_1} | {_2} | {km} |
mm |
cm |
m |
km |
|
{mm} |
{mm} ÷10 {cm} |
{cm} ÷100 {m} |
{m} ÷1000 {km} |
| mm | cm | m | km |
| {mm} | {_0} | {_1} | {_2} |
{A} - {B} = {var} |
||
{A} + {b} |
Rewrite the subtraction as
|
|
{=A-B} |
Follow the rules of additionThe signs are the same, find the sumThe sum of {a} and {b} = {=abs(a+b)} Keep the sign.The answer is {signAns} |
| {#a} |
| {#a} |
| {#a} |
{A} - {B} = {var} |
||
{A} + {b} |
Rewrite the subtraction as
|
|
{=A-B} |
Follow the rules of additionThe signs are different, take the differenceThe difference of {a} and {b} = {=abs(a-b)} Use the sign of the larger number.{larger} is larger, therefore the answer is {signAns} |
| {#a} |
{A} - {B} = {var} |
||
{A} + {b} |
Rewrite the subtraction as
|
|
{=A-B} |
Follow the rules of additionThe signs are the same, find the sumThe sum of {a} and {b} = {=abs(a+b)} Keep the sign.The answer is {signAns} |
| {#a} |
| {#a} |
| {#a} |
ab - {c}({d} - {e}) + c |
Substitute the values of the variable:a={a}, b={b}, and c = {f} |
||
{a}({b}) - {c}({d} - {e}) + {f} |
Parentheses: {d} - {e} |
 |
|
{a}({b}) - {c}({=d-e}) + {f} |
|
 |
|
{=a*b} - {c}({=d-e}) + {f} |
Multiply {c}({=d-e}) |
|
|
{=a*b} - {=c*(d-e)} + {f} |
Subtract {=a/b} - {=c*(d-e)} |
 |
|
{=a*b-(c*(d-e))} + {f} |
Add {=a*b-(c*(d-e))} + {f} |
|
|
{ans} |
{question} |
|||
|
1. Substitute {variable} = {varNum} | ||
{=number-varNum} |
2. Subtract |
{question} |
|||
{number}({varNum}) |
1. Substitute {variable} = {varNum}Place {varNum} in parentheses to show it is multiplication | ||
{=varNum*number} |
2. Multiply |
{question} |
1. Substitute {=variable[1]} = {number} |
|
|
2. Multiply {factor} x {number} |
|
{=factor*number} + {varNum} |
3. Add |
|
{=ans} |
{question} |
1. Substitute {=variable[1]} = {number} |
|
|
2. Multiply {factor} x {number} |
|
{=factor*number} - {varNum} |
3. Subtract |
|
{=ans} |
{question} |
|||
{varNum} + {number} |
1. Substitute {variable} = {varNum} | ||
{=varNum+number} |
2. Add |
{question} |
|||
{varNum} - {number} |
1. Substitute {variable} = {varNum} | ||
{=varNum-number} |
2. Subtract |
{question} |
|||
{varNum} - {number} |
1. Substitute {variable} = {varNum} | ||
{=varNum-number} |
2. Subtract |
{question} |
|||
{varNum} - {number} |
1. Substitute {variable} = {varNum} | ||
{=varNum-number} |
2. Subtract |
{question} |
|||
{varNum} - {number} |
1. Substitute {variable} = {varNum} | ||
{=varNum-number} |
2. Subtract |
{question} |
|||
{varNum} - {number} |
1. Substitute {variable} = {varNum} | ||
{=varNum-number} |
2. Subtract |
{question} |
|||
{varNum} + {number} |
1. Substitute {=variable[0]} = {varNum},and {=variable[1]} = {number} | ||
{=varNum+number} |
2. Add |
{question} |
|
|
|
|
1. Substitute {=variable[0]} = {varNum}, and {=variable[1]} = {number} |
||
{=number-varNum} |
2. Subtract |
Classify the triangle by its sides.
<A: {=choice[angA]} angle.
]]><B = {b}
<B: {=choice[angB]} angle.
<C = {c}
<C: {=choice[angC]} angle.
| x + {s} + 90 = \({180}^\circ\) | The sum of the angles = \({180}^\circ\) | |
| x + {=s+90} = \({180}^\circ\) | Simplify on the left side of the equation | |
| - {=s+90} = - {=s+90} |
Subtract {=s+90} from both sides | |
| x = \({{t}}^\circ\) | \(\angle\) A = \({{t}}^\circ\) |
| x + {t} + 90 = \({180}^\circ\) | The sum of the angles = \({180}^\circ\) | |
| x + {=90+t} = 180 | Add {t} + 90 | |
| -{=90+t} -{=90+t} | Subtract -{=90+t} from both sides | |
| x = {s} | Simplify | |
| \(\angle\)A = {s}°. |