QuestionSolve the following equations for the indicated variable:
27. $I=P r t$ for $P$
28. $C=2 \pi r$ for $r$
29. $T=D+p m$ for $p$
30. $P=C+M C$ for $M$
31. $A=\frac{1}{2} h(a+b)$ for $a$
32. $A=\frac{1}{2} h(a+b)$ for $b$
33. $S=P+$ Prt for $r$
34. $S=P+$ Prt for $t$
35. $B=\frac{F}{S-V}$ for $S$
36. $S=\frac{C}{1-r}$ for $r$

Studdy Solution

STEP 1

Assumptions1. All the given formulas are correct. . We are asked to solve each formula for the specified variable.

STEP 2

For the first formula, $=Prt$ , we are asked to solve for $$. We can do this by dividing both sides of the equation by $rt$.
$= \frac{}{rt}$

STEP 3

For the second formula, $C=2\pi r$ , we are asked to solve for $r$ . We can do this by dividing both sides of the equation by $2\pi$ .
$r = \frac{C}{2\pi}$

STEP 4

For the third formula, $=D+pm$ , we are asked to solve for $p$ . We can do this by subtracting $$ from both sides of the equation and then dividing by $m$.
$p = \frac{T-D}{m}$

STEP 5

For the fourth formula, $=C+MC$ , we are asked to solve for $M$ . We can do this by subtracting $C$ from both sides of the equation and then dividing by $C$ .
$M = \frac{-C}{C}$

STEP 6

For the fifth formula, $A=\frac{1}{2}h(a+b)$ , we are asked to solve for $a$ . We can do this by multiplying both sides of the equation by $2$ , dividing by $h$ , and then subtracting $b$ .
$a = \frac{2A}{h} - b$

STEP 7

For the sixth formula, $A=\frac{1}{2}h(a+b)$ , we are asked to solve for $b$ . We can do this by multiplying both sides of the equation by $2$ , dividing by $h$ , and then subtracting $a$ .
$b = \frac{2A}{h} - a$

STEP 8

For the seventh formula, $=P+rt$ , we are asked to solve for $r$ . We can do this by subtracting $$ from both sides of the equation, dividing by $Pt$.
$r = \frac{-P}{Pt}$

STEP 9

For the eighth formula, $=P+rt$ , we are asked to solve for $t$ . We can do this by subtracting $$ from both sides of the equation, dividing by $Pr$.
$t = \frac{-P}{Pr}$

STEP 10

For the ninth formula, $B=\frac{}{-V}$ , we are asked to solve for $$. We can do this by multiplying both sides of the equation by $-V$, then adding $BV$ to both sides.
$= \frac{+BV}{B}$

STEP 11

For the tenth formula, $=\frac{C}{-r}$ , we are asked to solve for $r$ . We can do this by multiplying both sides of the equation by $-r$ , then subtracting $$ from both sides and dividing by $-$.
$r = - \frac{C}{}$

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Studdy Solution

STEP 1

Assumptions1. All the given formulas are correct. . We are asked to solve each formula for the specified variable.

STEP 2

For the first formula, $=Prt$ , we are asked to solve for $$. We can do this by dividing both sides of the equation by $rt$.
$= \frac{}{rt}$

STEP 3

For the second formula, $C=2\pi r$ , we are asked to solve for $r$ . We can do this by dividing both sides of the equation by $2\pi$ .
$r = \frac{C}{2\pi}$

STEP 4

For the third formula, $=D+pm$ , we are asked to solve for $p$ . We can do this by subtracting $$ from both sides of the equation and then dividing by $m$.
$p = \frac{T-D}{m}$

STEP 5

For the fourth formula, $=C+MC$ , we are asked to solve for $M$ . We can do this by subtracting $C$ from both sides of the equation and then dividing by $C$ .
$M = \frac{-C}{C}$

STEP 6

For the fifth formula, $A=\frac{1}{2}h(a+b)$ , we are asked to solve for $a$ . We can do this by multiplying both sides of the equation by $2$ , dividing by $h$ , and then subtracting $b$ .
$a = \frac{2A}{h} - b$

STEP 7

For the sixth formula, $A=\frac{1}{2}h(a+b)$ , we are asked to solve for $b$ . We can do this by multiplying both sides of the equation by $2$ , dividing by $h$ , and then subtracting $a$ .
$b = \frac{2A}{h} - a$

STEP 8

For the seventh formula, $=P+rt$ , we are asked to solve for $r$ . We can do this by subtracting $$ from both sides of the equation, dividing by $Pt$.
$r = \frac{-P}{Pt}$

STEP 9

For the eighth formula, $=P+rt$ , we are asked to solve for $t$ . We can do this by subtracting $$ from both sides of the equation, dividing by $Pr$.
$t = \frac{-P}{Pr}$

STEP 10

For the ninth formula, $B=\frac{}{-V}$ , we are asked to solve for $$. We can do this by multiplying both sides of the equation by $-V$, then adding $BV$ to both sides.
$= \frac{+BV}{B}$

STEP 11

For the tenth formula, $=\frac{C}{-r}$ , we are asked to solve for $r$ . We can do this by multiplying both sides of the equation by $-r$ , then subtracting $$ from both sides and dividing by $-$.
$r = - \frac{C}{}$

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Studdy Solution

STEP 1

Assumptions1. All the given formulas are correct. . We are asked to solve each formula for the specified variable.

STEP 2

For the first formula, =Prt=Prt=Prt, we are asked to solve for $$. We can do this by dividing both sides of the equation by $rt$.=rt = \frac{}{rt}=rt​

STEP 3

For the second formula, C=2πrC=2\pi rC=2πr, we are asked to solve for rrr. We can do this by dividing both sides of the equation by 2π2\pi2π.r=C2πr = \frac{C}{2\pi}r=2πC​

STEP 4

For the third formula, =D+pm=D+pm=D+pm, we are asked to solve for ppp. We can do this by subtracting $$ from both sides of the equation and then dividing by $m$.p=T−Dmp = \frac{T-D}{m}p=mT−D​

STEP 5

For the fourth formula, =C+MC=C+MC=C+MC, we are asked to solve for MMM. We can do this by subtracting CCC from both sides of the equation and then dividing by CCC.M=−CCM = \frac{-C}{C}M=C−C​

STEP 6

For the fifth formula, A=12h(a+b)A=\frac{1}{2}h(a+b)A=21​h(a+b), we are asked to solve for aaa. We can do this by multiplying both sides of the equation by 222, dividing by hhh, and then subtracting bbb.a=2Ah−ba = \frac{2A}{h} - ba=h2A​−b

STEP 7

For the sixth formula, A=12h(a+b)A=\frac{1}{2}h(a+b)A=21​h(a+b), we are asked to solve for bbb. We can do this by multiplying both sides of the equation by 222, dividing by hhh, and then subtracting aaa.b=2Ah−ab = \frac{2A}{h} - ab=h2A​−a

STEP 8

For the seventh formula, =P+rt=P+rt=P+rt, we are asked to solve for rrr. We can do this by subtracting $$ from both sides of the equation, dividing by $Pt$.r=−PPtr = \frac{-P}{Pt}r=Pt−P​

STEP 9

For the eighth formula, =P+rt=P+rt=P+rt, we are asked to solve for ttt. We can do this by subtracting $$ from both sides of the equation, dividing by $Pr$.t=−PPrt = \frac{-P}{Pr}t=Pr−P​

STEP 10

For the ninth formula, B=−VB=\frac{}{-V}B=−V​, we are asked to solve for $$. We can do this by multiplying both sides of the equation by $-V$, then adding $BV$ to both sides.=+BVB = \frac{+BV}{B}=B+BV​

STEP 11

For the tenth formula, =C−r=\frac{C}{-r}=−rC​, we are asked to solve for rrr. We can do this by multiplying both sides of the equation by −r-r−r, then subtracting $$ from both sides and dividing by $-$.r=−Cr = - \frac{C}{}r=−C​

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For the first formula, $=Prt$ , we are asked to solve for $$. We can do this by dividing both sides of the equation by $rt$.
$= \frac{}{rt}$

For the second formula, $C=2\pi r$ , we are asked to solve for $r$ . We can do this by dividing both sides of the equation by $2\pi$ .
$r = \frac{C}{2\pi}$

For the third formula, $=D+pm$ , we are asked to solve for $p$ . We can do this by subtracting $$ from both sides of the equation and then dividing by $m$.
$p = \frac{T-D}{m}$

For the fourth formula, $=C+MC$ , we are asked to solve for $M$ . We can do this by subtracting $C$ from both sides of the equation and then dividing by $C$ .
$M = \frac{-C}{C}$

For the fifth formula, $A=\frac{1}{2}h(a+b)$ , we are asked to solve for $a$ . We can do this by multiplying both sides of the equation by $2$ , dividing by $h$ , and then subtracting $b$ .
$a = \frac{2A}{h} - b$

For the sixth formula, $A=\frac{1}{2}h(a+b)$ , we are asked to solve for $b$ . We can do this by multiplying both sides of the equation by $2$ , dividing by $h$ , and then subtracting $a$ .
$b = \frac{2A}{h} - a$

For the seventh formula, $=P+rt$ , we are asked to solve for $r$ . We can do this by subtracting $$ from both sides of the equation, dividing by $Pt$.
$r = \frac{-P}{Pt}$

For the eighth formula, $=P+rt$ , we are asked to solve for $t$ . We can do this by subtracting $$ from both sides of the equation, dividing by $Pr$.
$t = \frac{-P}{Pr}$

For the ninth formula, $B=\frac{}{-V}$ , we are asked to solve for $$. We can do this by multiplying both sides of the equation by $-V$, then adding $BV$ to both sides.
$= \frac{+BV}{B}$

For the tenth formula, $=\frac{C}{-r}$ , we are asked to solve for $r$ . We can do this by multiplying both sides of the equation by $-r$ , then subtracting $$ from both sides and dividing by $-$.
$r = - \frac{C}{}$