Creating Equations: Rearranging Formulas to Highlight Quantities (CCSS.A-CED.4)
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Common Core High School Algebra › Creating Equations: Rearranging Formulas to Highlight Quantities (CCSS.A-CED.4)
Which is the correct formula for $R$ in Ohm's law $V=IR$?
$R=VI$
$R=\frac{I}{V}$
$R=\frac{V}{I}$
$R=V-I$
Explanation
Starting with $V=IR$, divide both sides by $I$ to isolate $R$: $\dfrac{V}{I}=R$. This matters because it lets you find resistance from measured voltage and current.
Which is the correct formula for the radius $r$ in the circle area formula $A=\pi r^2$?
$r=\sqrt{\dfrac{A}{\pi}}$
$r=\dfrac{A}{\pi}$
$r=\dfrac{\pi}{A}$
$r=\sqrt{A},\pi$
Explanation
From $A=\pi r^2$, divide by $\pi$: $\dfrac{A}{\pi}=r^2$. Then take the square root: $r=\sqrt{\dfrac{A}{\pi}}$. This rearrangement lets you find the radius when the area is known.
The perimeter of a rectangle is given by $P=2\ell+2w$. Which formula correctly isolates $\ell$ in terms of $P$ and $w$?
$\ell=\dfrac{P}{2}+w$
$\ell=\dfrac{P}{2w}-1$
$\ell=P-2w$
$\ell=\dfrac{P-2w}{2}$
Explanation
Subtract $2w$ from both sides: $P-2w=2\ell$. Then divide both sides by $2$: $\ell=\dfrac{P-2w}{2}$. This expresses length in terms of perimeter and width.
Using $F=ma$, what is the value of $m$ when $F=24\ \text{N}$ and $a=3\ \text{m/s}^2$?
72 kg
8 kg
21 kg
0.125 kg
Explanation
Solve for mass by dividing both sides by acceleration: $m=\dfrac{F}{a}=\dfrac{24\ \text{N}}{3\ \text{m/s}^2}=8\ \text{kg}$. Rearranging shows how force, mass, and acceleration relate.
Simple interest is given by $I=Prt$. If $I=240$, $P=800$, and $r=0.05$, what is the value of $t$?
6 years
0.6 years
0.015 years
4 years
Explanation
Isolate time: $t=\dfrac{I}{Pr}=\dfrac{240}{800\cdot 0.05}=\dfrac{240}{40}=6$. Rearranging allows you to compute how long money is invested at a given rate.
Ohm's law relates voltage $V$, current $I$, and resistance $R$ by $V = IR$. Which is the correct formula for resistance $R$ in terms of $V$ and $I$?
$R = \dfrac{V}{I}$
$R = \dfrac{I}{V}$
$R = VI$
$R = V - I$
Explanation
Starting with $V = IR$, divide both sides by $I$ to isolate $R$: $\dfrac{V}{I} = R$. In context, resistance is the ratio of voltage to current, so $R = \dfrac{V}{I}$. The distractors either invert the ratio, multiply instead of divide, or subtract, which do not follow valid equation-solving steps.
The area of a circle is given by $A = \pi r^2$. Which is the correct formula for the radius $r$ in terms of $A$ and $\pi$?
$r = \dfrac{A}{\pi}$
$r = \sqrt{\dfrac{A}{\pi}}$
$r = \dfrac{A^2}{\pi}$
$r = \dfrac{\pi}{A}$
Explanation
From $A = \pi r^2$, divide both sides by $\pi$: $\dfrac{A}{\pi} = r^2$. Then take the square root of both sides to solve for $r$: $r = \sqrt{\dfrac{A}{\pi}}$. In context, radius must be nonnegative, so we take the positive root.
A cyclist travels at a constant speed $r = 52$ miles/hour for a trip of $d = 156$ miles. Using $d = rt$, what is the value of $t$?
0.33 hours
52 hours
3 hours
208 hours
Explanation
Solve $d = rt$ for $t$ by dividing both sides by $r$: $t = \dfrac{d}{r}$. Substitute $d = 156$ and $r = 52$: $t = \dfrac{156}{52} = 3$ hours. Distractors come from inverting the ratio, using $r$ itself, or multiplying $d$ and $r$.
Simple interest is modeled by $I = Prt$, where $I$ is interest, $P$ principal, $r$ annual rate, and $t$ time in years. Which is the correct formula for $t$?
$t = \dfrac{IP}{r}$
$t = \dfrac{Ir}{P}$
$t = \dfrac{P}{Ir}$
$t = \dfrac{I}{Pr}$
Explanation
From $I = Prt$, divide both sides by $Pr$ to isolate $t$: $t = \dfrac{I}{Pr}$. This shows time is interest divided by the product of principal and rate. The distractors misplace factors or invert the division, which does not keep the equation balanced.
Density is given by $D = \dfrac{m}{V}$, where $m$ is mass and $V$ is volume. If $m = 4.5$ kg and $D = 0.9$ kg/L, what is the value of $V$?
5 L
0.2 L
4.5 L
4.05 L
Explanation
Rearrange $D = \dfrac{m}{V}$ to isolate $V$: multiply both sides by $V$ to get $DV = m$, then divide by $D$: $V = \dfrac{m}{D}$. Substitute $m = 4.5$ and $D = 0.9$: $V = \dfrac{4.5}{0.9} = 5$ L. Distractors come from inverting the fraction, forgetting to divide, or subtracting instead of dividing.