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Common Core: High School - Algebra : Creating Equations✭

Study concepts, example questions & explanations for Common Core: High School - Algebra

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All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

The equation for a circle centered at (0,0) is,

$\displaystyle x^2+y^2=r^2$

Solve the equation for $\displaystyle r$ .

Possible Answers:

$r=\sqrt{x^2+y^2}$ $\displaystyle r=\sqrt{x^2+y^2}$

$r=\sqrt{x^2-y^2}$ $\displaystyle r=\sqrt{x^2-y^2}$

r=x+y $\displaystyle r=x+y$

$\displaystyle r=x^2+y^2$

r=x-y $\displaystyle r=x-y$

Correct answer:

$r=\sqrt{x^2+y^2}$ $\displaystyle r=\sqrt{x^2+y^2}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle r$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

To isolate $\displaystyle r$ take the square root of each side.

$\displaystyle x^2+y^2=r^2$

$\sqrt{x^2+y^2}=\sqrt{r^2}$ $\displaystyle \sqrt{x^2+y^2}=\sqrt{r^2}$

$\sqrt{x^2+y^2}=r$ $\displaystyle \sqrt{x^2+y^2}=r$

Therefore, the equation solved for $\displaystyle r$ is

$r=\sqrt{x^2+y^2}$ $\displaystyle r=\sqrt{x^2+y^2}$

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Example Question #4 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

The equation for a circle centered at (0,0) is,

$\displaystyle x^2+y^2=r^2$

Solve the equation for $\displaystyle x$ .

Possible Answers:

$x=\sqrt{r^2+y^2}$ $\displaystyle x=\sqrt{r^2+y^2}$

x=r-y $\displaystyle x=r-y$

$x=\sqrt{-r^2-y^2}$ $\displaystyle x=\sqrt{-r^2-y^2}$

$x=\sqrt{r^2-y^2}$ $\displaystyle x=\sqrt{r^2-y^2}$

x=r+y $\displaystyle x=r+y$

Correct answer:

$x=\sqrt{r^2-y^2}$ $\displaystyle x=\sqrt{r^2-y^2}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle x$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

To isolate $\displaystyle x$ subtract y^2 $\displaystyle y^2$ from both sides and then take the square root.

$\displaystyle x^2+y^2=r^2$

$\displaystyle x^2+y^2-y^2=r^2-y^2$

$\displaystyle x^2=r^2-y^2$

$\sqrt{x^2}=\sqrt{r^2-y^2}$ $\displaystyle \sqrt{x^2}=\sqrt{r^2-y^2}$

Therefore, the equation solved for $\displaystyle x$ is

$x=\sqrt{r^2-y^2}$ $\displaystyle x=\sqrt{r^2-y^2}$

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Example Question #1 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

The equation for a circle centered at (0,0) is,

$\displaystyle x^2+y^2=r^2$

Solve the equation for $\displaystyle y$ .

Possible Answers:

y=r-x $\displaystyle y=r-x$

$y=\sqrt{-r^2-x^2}$ $\displaystyle y=\sqrt{-r^2-x^2}$

$y=\sqrt{r^2+x^2}$ $\displaystyle y=\sqrt{r^2+x^2}$

y=r+x $\displaystyle y=r+x$

$y=\sqrt{r^2-x^2}$ $\displaystyle y=\sqrt{r^2-x^2}$

Correct answer:

$y=\sqrt{r^2-x^2}$ $\displaystyle y=\sqrt{r^2-x^2}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle y$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

To isolate $\displaystyle y$ subtract x^2 $\displaystyle x^2$ from both sides and then take the square root.

$\displaystyle x^2+y^2=r^2$

$\displaystyle x^2+y^2-x^2=r^2-x^2$

$\displaystyle y^2=r^2-x^2$

$\sqrt{y^2}=\sqrt{r^2-x^2}$ $\displaystyle \sqrt{y^2}=\sqrt{r^2-x^2}$

Therefore, the equation solved for $\displaystyle y$ is

$y=\sqrt{r^2-x^2}$ $\displaystyle y=\sqrt{r^2-x^2}$

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Example Question #1 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

To calculate simple interest the following formula is used.

I=PRT $\displaystyle I=PRT$

Solve the equation for $\displaystyle P$ .

Possible Answers:

$P=\frac{I}{RT}$ $\displaystyle P=\frac{I}{RT}$

$P=\frac{I}{R}-T$ $\displaystyle P=\frac{I}{R}-T$

P=IRT $\displaystyle P=IRT$

P=I-RT $\displaystyle P=I-RT$

$P=\frac{I}{T}-R$ $\displaystyle P=\frac{I}{T}-R$

Correct answer:

$P=\frac{I}{RT}$ $\displaystyle P=\frac{I}{RT}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle P$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since $\displaystyle P$ is being multiplied by $\displaystyle RT$ , divide by $\displaystyle RT$ on both sides.

I=PRT $\displaystyle I=PRT$

$\frac{I}{RT}=\frac{PRT}{RT}$ $\displaystyle \frac{I}{RT}=\frac{PRT}{RT}$

$\frac{I}{RT}=P$ $\displaystyle \frac{I}{RT}=P$

Therefore, the equation solved for $\displaystyle P$ is,

$P=\frac{I}{RT}$ $\displaystyle P=\frac{I}{RT}$

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Example Question #9 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

To calculate simple interest the following formula is used.

I=PRT $\displaystyle I=PRT$

Solve the equation for $\displaystyle R$ .

Possible Answers:

$R=\frac{I}{PT}$ $\displaystyle R=\frac{I}{PT}$

$R=\frac{I}{P}-T$ $\displaystyle R=\frac{I}{P}-T$

$R=\frac{I}{T}-P$ $\displaystyle R=\frac{I}{T}-P$

R=IPT $\displaystyle R=IPT$

$R=-\frac{I}{PT}$ $\displaystyle R=-\frac{I}{PT}$

Correct answer:

$R=\frac{I}{PT}$ $\displaystyle R=\frac{I}{PT}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle R$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since $\displaystyle R$ is being multiplied by $\displaystyle PT$ , divide by $\displaystyle PT$ on both sides.

I=PRT $\displaystyle I=PRT$

$\frac{I}{PT}=\frac{PRT}{PT}$ $\displaystyle \frac{I}{PT}=\frac{PRT}{PT}$

$\frac{I}{PT}=R$ $\displaystyle \frac{I}{PT}=R$

Therefore, the equation solved for $\displaystyle R$ is,

$R=\frac{I}{PT}$ $\displaystyle R=\frac{I}{PT}$

Report an Error

Example Question #10 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

To calculate simple interest the following formula is used.

I=PRT $\displaystyle I=PRT$

Solve the equation for $\displaystyle T$ .

Possible Answers:

$T=\frac{I}{R}-P$ $\displaystyle T=\frac{I}{R}-P$

$T=-\frac{I}{PR}$ $\displaystyle T=-\frac{I}{PR}$

$T=\frac{I}{P}-R$ $\displaystyle T=\frac{I}{P}-R$

T=I-PR $\displaystyle T=I-PR$

$T=\frac{I}{PR}$ $\displaystyle T=\frac{I}{PR}$

Correct answer:

$T=\frac{I}{PR}$ $\displaystyle T=\frac{I}{PR}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle T$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since $\displaystyle T$ is being multiplied by $\displaystyle PR$ , divide by $\displaystyle PR$ on both sides.

I=PRT $\displaystyle I=PRT$

$\frac{I}{PR}=\frac{PRT}{PR}$ $\displaystyle \frac{I}{PR}=\frac{PRT}{PR}$

$\frac{I}{PR}=T$ $\displaystyle \frac{I}{PR}=T$

Therefore, the equation solved for $\displaystyle T$ is,

$T=\frac{I}{PR}$ $\displaystyle T=\frac{I}{PR}$

Report an Error

Example Question #451 : High School: Algebra

To calculate the volume of a rectangle the following formula is used.

V=lwh $\displaystyle V=lwh$

Solve the equation for $\displaystyle l$ .

Possible Answers:

l=V-wh $\displaystyle l=V-wh$

$l=\frac{V}{h}-w$ $\displaystyle l=\frac{V}{h}-w$

$l=-\frac{V}{wh}$ $\displaystyle l=-\frac{V}{wh}$

$l=\frac{V}{w}-h$ $\displaystyle l=\frac{V}{w}-h$

$l=\frac{V}{wh}$ $\displaystyle l=\frac{V}{wh}$

Correct answer:

$l=\frac{V}{wh}$ $\displaystyle l=\frac{V}{wh}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle l$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since $\displaystyle l$ is being multiplied by $\displaystyle wh$ , divide by $\displaystyle wh$ on both sides.

V=lwh $\displaystyle V=lwh$

$\frac{V}{wh}=\frac{lwh}{wh}$ $\displaystyle \frac{V}{wh}=\frac{lwh}{wh}$

$\frac{V}{wh}=l$ $\displaystyle \frac{V}{wh}=l$

Therefore, the equation solved for $\displaystyle l$ is,

$l=\frac{V}{wh}$ $\displaystyle l=\frac{V}{wh}$

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Example Question #41 : Creating Equations✭

To calculate the volume of a rectangle the following formula is used.

V=lwh $\displaystyle V=lwh$

Solve the equation for $\displaystyle h$ .

Possible Answers:

h=V-wl $\displaystyle h=V-wl$

$h=\frac{V}{wl}$ $\displaystyle h=\frac{V}{wl}$

$h=\frac{V}{w}-l$ $\displaystyle h=\frac{V}{w}-l$

$h=-\frac{V}{wl}$ $\displaystyle h=-\frac{V}{wl}$

$h=\frac{V}{l}-w$ $\displaystyle h=\frac{V}{l}-w$

Correct answer:

$h=\frac{V}{wl}$ $\displaystyle h=\frac{V}{wl}$

Explanation:

This question is asking to rearrange this formula to highlight a quantity of interest, specifically $\displaystyle h$ .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since $\displaystyle h$ is being multiplied by $\displaystyle wl$ , divide by $\displaystyle wl$ on both sides.

V=lwh $\displaystyle V=lwh$

$\frac{V}{wl}=\frac{lwh}{wl}$ $\displaystyle \frac{V}{wl}=\frac{lwh}{wl}$

$\frac{V}{wl}=h$ $\displaystyle \frac{V}{wl}=h$

Therefore, the equation solved for $\displaystyle h$ is,

$h=\frac{V}{wl}$ $\displaystyle h=\frac{V}{wl}$

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All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

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Common Core: High School - Algebra : Creating Equations✭

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #3 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

Example Question #4 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

Example Question #1 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

Example Question #1 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

Example Question #9 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

Example Question #10 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

Example Question #451 : High School: Algebra

Example Question #41 : Creating Equations✭

All Common Core: High School - Algebra Resources

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