Question 1

Solve the equation for the specified variable.
-$\frac { 1 } { R } = \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 2 } } \quad$ for $R _ { 1 }$
A) $R _ { 1 } = \frac { R _ { 2 } } { R _ { 2 } - R }$
B) $R _ { 1 } = \frac { R _ { 2 } - R } { R _ { 2 } }$
C) $R _ { 1 } = \frac { R R _ { 2 } } { R _ { 2 } - R }$
D) $R _ { 1 } = \frac { R } { R _ { 2 } - R }$

Accepted Answer

To solve for $R_1$, first combine the fractions on the right side of the equation by finding a common denominator, which gives $\frac{1}{R} = \frac{R_1 + R_2}{R_1R_2}$. Then, isolate $R_1$ by cross-multiplying and rearranging, leading to $R_1 = \frac{RR_2}{R_2 - R}$.

Question 2

Solve.
-If three times a number added to 7 is divided by the number plus 9, the result is four thirds. Find the number. 
A) $- \frac { 21 } { 2 }$
B) $- \frac { 1 } { 9 }$
C) 3
D) $- \frac { 27 } { 5 }$

Accepted Answer

The answer of Solve.
-If three times a number added to...

Question 3

Solve.
-One number is 5 times larger than another. The sum of their reciprocals is $\frac { 2 } { 5 }$. Determine the two numbers.
A) 15 and 75
B) 3 and 15
C) $\frac { 1 } { 3 }$ and $\frac { 5 } { 3 }$
D) 5 and 25

Accepted Answer

Let the smaller number be $x$ and the larger number be $5x$. The sum of their reciprocals is $\frac{1}{x} + \frac{1}{5x} = \frac{2}{5}$. Solving this equation gives $x = 3$ and $5x = 15$, which matches option B.

Question 4

Solve.
-Two times the reciprocal of a number equals 32 times the reciprocal of 40. Find the number. 
A) 10
B) $\frac { 5 } { 2 }$
C) $\frac { 8 } { 5 }$
D) 80

Accepted Answer

The answer of Solve.
-Two times the reciprocal of a number...

Question 5

Solve the equation.
-$$\left( \frac { 7 } { x + 3 } \right) ^ { 2 } + \left( \frac { 7 } { x + 3 } \right) - 6 = 0$$
A) $\left\{ \frac { 16 } { 3 } , \frac { 1 } { 2 } \right\}$
B) $\left\{ - \frac { 16 } { 3 } , \frac { 1 } { 2 } \right\}$
C) $\left\{ \frac { 16 } { 3 } , - \frac { 1 } { 2 } \right\}$
D) $\left\{ - \frac { 16 } { 3 } , - \frac { 1 } { 2 } \right\}$

Accepted Answer

The answer of Solve the equation.
-\[\left( \frac { 7 }...

Question 6

Solve the problem.
-The average cost of producing $x$ specialty phones is given by the function $f ( x ) = 15 + \frac { 3600 } { x }$. Find the number of phones that must be produced for the average cost to be $\$ 24$.
A) 360 phones
B) 40 phones
C) 440 phones
D) 400 phones

Accepted Answer

To find the number of phones, set $f(x) = 24$: $15 + \frac{3600}{x} = 24$. Solving for $x$, we get $ \frac{3600}{x} = 9 $, so $x = \frac{3600}{9} = 400$.

Question 7

Solve the equation for the specified variable.
-$\frac { \mathrm { P } _ { 1 } \mathrm {~V} _ { 1 } } { \mathrm {~T} _ { 1 } } = \frac { \mathrm { P } _ { 2 } \mathrm {~V} _ { 2 } } { \mathrm {~T} _ { 2 } } \quad$ for $\mathrm { P } _ { 1 }$
A) $\mathrm { P } _ { 1 } = \frac { \mathrm { V } _ { 1 } \mathrm {~V} _ { 2 } \mathrm {~T} _ { 1 } } { \mathrm { P } _ { 2 } \mathrm {~T} _ { 2 } }$
B) $\mathrm { P } _ { 1 } = \frac { \mathrm { P } _ { 2 } \mathrm {~V} _ { 2 } \mathrm {~T} _ { 1 } } { \mathrm {~V} _ { 1 } \mathrm {~T} _ { 2 } }$
C) $\mathrm { P } _ { 1 } = \frac { \mathrm { P } _ { 2 } \mathrm {~V} _ { 1 } \mathrm {~T} _ { 1 } } { \mathrm {~V} _ { 2 } \mathrm {~T} _ { 2 } }$
D) $\mathrm { P } _ { 1 } = \frac { \mathrm { P } _ { 2 } \mathrm {~V} _ { 2 } \mathrm {~T} _ { 2 } } { \mathrm {~V} _ { 1 } \mathrm {~T} _ { 1 } }$

Accepted Answer

The answer of Solve the equation for the specified variable.
-\(\frac...

Question 8

Solve.
-Five divided by the difference of a number and 9 equals the quotient of 10 and the sum of the number and 11. Find the number. 
A) $- 7$
B) 4
C) $\frac { 29 } { 3 }$
D) 29

Accepted Answer

The answer of Solve.
-Five divided by the difference of a...

Question 9

Solve the equation for the specified variable.
-$\mathrm { f } = \frac { \mathrm { f } _ { 1 } \mathrm { f } _ { 2 } } { \mathrm { f } _ { 1 } + \mathrm { f } _ { 2 } } \quad$ for $\mathrm { f } _ { 2 }$
A) $f _ { 2 } = \frac { f - f _ { 1 } } { f _ { 1 } - f }$
B) $f _ { 2 } = \frac { f f _ { 1 } } { f _ { 1 } - f }$
C) $f _ { 2 } = \frac { f f _ { 1 } } { f _ { 1 } + f }$
D) $f _ { 2 } = - 1$

Accepted Answer

To solve for $f_2$, start with the given equation $f = \frac{f_1 f_2}{f_1 + f_2}$. Multiply both sides by $f_1 + f_2$ to get $f(f_1 + f_2) = f_1 f_2$. Then, isolate $f_2$ by dividing both sides by $f_1 - f$, resulting in $f_2 = \frac{f f_1}{f_1 - f}$.

Question 10

Solve the equation for the specified variable.
-$A = \frac { h ( a + b ) } { 2 } \quad$ for $a$
A) $a = \frac { A - h b } { 2 h }$
B) $\mathrm { a } = \frac { 2 \mathrm {~A} - \mathrm { hb } } { \mathrm { h } }$
C) $\mathrm { a } = \frac { h b - 2 \mathrm {~A} } { \mathrm {~h} }$
D) $a = \frac { 2 \mathrm { Ab } - \mathrm { h } } { \mathrm { h } }$

Accepted Answer

To solve for $a$, first multiply both sides by 2 to get rid of the fraction: $2A = h(a + b)$. Then, isolate $a$ by subtracting $hb$ from both sides and finally divide by $h$: $a = \frac{2A - hb}{h}$.

Question 11

Solve.
-A recent advertisement claimed that 2 out of every 5 doctors recommend a certain herbal supplement to increase energy levels. If a local hospital employs 180 doctors, how many doctors would you expect to
Recommend the supplement? (Round to the nearest whole number, if necessary.)
A) 72 doctors
B) 36 doctors
C) 450 doctors
D) 1800 doctors

Accepted Answer

The advertisement claims 2 out of every 5 doctors recommend the supplement. Therefore, to find the expected number of doctors out of 180, you calculate (2/5) * 180 = 72 doctors.

Question 12

Solve the equation.
-$$x ^ { - 2 } - 2 x ^ { - 1 } - 15 = 0$$
A) $\left\{ \frac { 1 } { 5 } , - \frac { 1 } { 3 } \right\}$
B) $\{ - 15,1 \}$
C) $\left\{ - \frac { 1 } { 5 } , \frac { 1 } { 3 } \right\}$
D) $\{ - 5 , - 3 \}$

Accepted Answer

To solve the equation $x^{-2} - 2x^{-1} - 15 = 0$, let $y = x^{-1}$, then the equation becomes $y^2 - 2y - 15 = 0$. Factoring this quadratic equation gives $(y-5)(y+3) = 0$, so $y = 5$ or $y = -3$. Reverting back to $x$, we get $x = \frac{1}{5}$ or $x = -\frac{1}{3}$, which corresponds to choice A.

Question 13

Solve.
-Five divided by the sum of a number and 10, minus the quotient of 3 and the difference of the number and 10 is equal to 6 times the reciprocal of the difference of the number squared and 100. What is the number? 
A) 43
B) 13
C) $- 37$
D) $- \frac { 37 } { 4 }$

Accepted Answer

The correct answer is A because 43 is the only number that satisfies the equation.

Question 14

Solve the equation for the specified variable.
-$W = \frac { \mathrm { P } } { 2 } - \mathrm { L } \quad$ for $\mathrm { P }$
A) $P = 2 W - 2 L$
B) $\mathrm { P } = 2 \mathrm {~W} + \mathrm { L }$
C) $P = 2 W + 2 L$
D) $P = 2 W L$

Accepted Answer

The answer of Solve the equation for the specified variable.
-\(W...

Question 15

Solve.
-The denominator of the fraction $\frac { 4 } { 13 }$ is increased by an amount so that the value of the resulting fraction is $\frac { 2 } { 11 }$. Determine the amount by which the denominator was increased.
A) 9
B) 35
C) $\frac { 18 } { 11 }$
D) 10

Accepted Answer

The original fraction is $\frac{4}{13}$. To find the amount by which the denominator was increased to get the new fraction $\frac{2}{11}$, we set up the equation $\frac{4}{13+x} = \frac{2}{11}$, where $x$ is the amount added to the denominator. Cross-multiplying gives $44 = 26 + 2x$, leading to $18 = 2x$, so $x = 9$.

Question 16

Solve.
-The ratio of the weight of an object on Earth to an object on Planet X is 2 to 7. If a person weighs 220 pounds on Earth, find his weight on Planet X. (Round to the nearest whole number, if necessary.)
A) 16 pounds
B) 770 pounds
C) 63 pounds
D) 3080 pounds

Accepted Answer

The ratio of the weight of an object on Earth to Planet X is 2:7. To find the weight on Planet X, you multiply the person's Earth weight by 7/2. So, $220 \times \frac{7}{2} = 770$ pounds.

Question 17

Solve the equation for the specified variable.
-$P = \frac { A } { 1 + r t } \quad$ for $r$
A) $r = P - A t$
B) $r = \frac { P - A } { 1 + t }$
C) $r = \frac { A - P } { P t }$
D) $r = \frac { P - 1 } { A t }$

Accepted Answer

To solve for $r$, we start with the given equation $P = \frac { A } { 1 + r t }$. We want to isolate $r$, so we first multiply both sides by $1 + rt$ to get rid of the denominator, leading to $P(1 + rt) = A$. Expanding the left side gives $P + Prt = A$, and then we isolate $rt$ by subtracting $P$ from both sides, resulting in $Prt = A - P$. Finally, we divide both sides by $Pt$ to solve for $r$, yielding $r = \frac { A - P } { P t }$.

Question 18

Solve.
-To estimate the number of people in Springfield, population 10,000, who have a swimming pool in their backyard, 250 people were interviewed. Of those polled, 60 had a swimming pool. How many people in the
City might one expect to have a swimming pool? (Round to the nearest whole number, if necessary.)
A) 2 people
B) 41,667 people
C) 2400 people
D) 240 people

Accepted Answer

The proportion of people with a swimming pool in the sample is 60/250. Applying this proportion to the entire population of 10,000 gives (60/250) * 10,000 = 2400 people.

Question 19

Solve the equation for the specified variable.
-$$F = \frac { - G M m } { r ^ { 2 } } \text { for } G$$
A) $G = \frac { - F M m } { r ^ { 2 } }$
B) $G = \frac { - \mathrm { Fr } ^ { 2 } } { \mathrm { Mm } }$
C) $G = \frac { F ^ { 2 } } { M m }$
D) $G = - \mathrm { Fr } ^ { 2 } - \mathrm { Mm }$

Accepted Answer

To solve for $G$, you need to isolate $G$ on one side of the equation. Starting with $F = \frac{-GMm}{r^2}$, multiply both sides by $-\frac{r^2}{Mm}$ to get $G = \frac{-Fr^2}{Mm}$.

Question 20

Solve the equation.
-$80 x ^ { - 2 } - 18 x ^ { - 1 } + 1 = 0$
A) $\left\{ \frac { 1 } { 8 } , \frac { 1 } { 10 } \right\}$
B) $\{ 8 , - 10 \}$
C) $\{ 8,10 \}$
D) $\{ - 8 , - 10 \}$

Accepted Answer

The answer of Solve the equation.
-\(80 x ^ { -...

Quiz 6: Rational Expressions

Solve the equation for the specified variable. - $\frac { 1 } { R } = \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 2 } } \quad$ for $R _ { 1 }$

Solve. -If three times a number added to 7 is divided by the number plus 9, the result is four thirds. Find the number.

Solve. -One number is 5 times larger than another. The sum of their reciprocals is $\frac { 2 } { 5 }$ . Determine the two numbers.

Solve. -Two times the reciprocal of a number equals 32 times the reciprocal of 40. Find the number.

Solve the equation. - $\left( \frac { 7 } { x + 3 } \right) ^ { 2 } + \left( \frac { 7 } { x + 3 } \right) - 6 = 0$

Solve the problem. -The average cost of producing $x$ specialty phones is given by the function $f ( x ) = 15 + \frac { 3600 } { x }$ . Find the number of phones that must be produced for the average cost to be $\$ 24$ .

Solve the equation for the specified variable. - $\frac { \mathrm { P } _ { 1 } \mathrm {~V} _ { 1 } } { \mathrm {~T} _ { 1 } } = \frac { \mathrm { P } _ { 2 } \mathrm {~V} _ { 2 } } { \mathrm {~T} _ { 2 } } \quad$ for $\mathrm { P } _ { 1 }$

Solve. -Five divided by the difference of a number and 9 equals the quotient of 10 and the sum of the number and 11. Find the number.

Solve the equation for the specified variable. - $\mathrm { f } = \frac { \mathrm { f } _ { 1 } \mathrm { f } _ { 2 } } { \mathrm { f } _ { 1 } + \mathrm { f } _ { 2 } } \quad$ for $\mathrm { f } _ { 2 }$

Solve the equation for the specified variable. - $A = \frac { h ( a + b ) } { 2 } \quad$ for $a$

Solve. -A recent advertisement claimed that 2 out of every 5 doctors recommend a certain herbal supplement to increase energy levels. If a local hospital employs 180 doctors, how many doctors would you expect to Recommend the supplement? (Round to the nearest whole number, if necessary.)

Solve the equation. - $x ^ { - 2 } - 2 x ^ { - 1 } - 15 = 0$

Solve. -Five divided by the sum of a number and 10, minus the quotient of 3 and the difference of the number and 10 is equal to 6 times the reciprocal of the difference of the number squared and 100. What is the number?

Solve the equation for the specified variable. - $W = \frac { \mathrm { P } } { 2 } - \mathrm { L } \quad$ for $\mathrm { P }$

Solve. -The denominator of the fraction $\frac { 4 } { 13 }$ is increased by an amount so that the value of the resulting fraction is $\frac { 2 } { 11 }$ . Determine the amount by which the denominator was increased.

Solve. -The ratio of the weight of an object on Earth to an object on Planet X is 2 to 7. If a person weighs 220 pounds on Earth, find his weight on Planet X. (Round to the nearest whole number, if necessary.)

Solve the equation for the specified variable. - $P = \frac { A } { 1 + r t } \quad$ for $r$

Solve the equation for the specified variable. - $F = \frac { - G M m } { r ^ { 2 } } \text { for } G$

Solve the equation. - $80 x ^ { - 2 } - 18 x ^ { - 1 } + 1 = 0$

Quiz 6: Rational Expressions

Solve the equation for the specified variable. - 1R=1R1+1R2\frac { 1 } { R } = \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 2 } } \quadR1​=R1​1​+R2​1​ for R1R _ { 1 }R1​

Solve. -If three times a number added to 7 is divided by the number plus 9, the result is four thirds. Find the number.

Solve. -One number is 5 times larger than another. The sum of their reciprocals is 25\frac { 2 } { 5 }52​ . Determine the two numbers.

Solve. -Two times the reciprocal of a number equals 32 times the reciprocal of 40. Find the number.

Solve the equation. - (7x+3) 2+(7x+3) −6=0\left( \frac { 7 } { x + 3 } \right) ^ { 2 } + \left( \frac { 7 } { x + 3 } \right) - 6 = 0(x+37​) 2+(x+37​) −6=0

Solve the problem. -The average cost of producing xxx specialty phones is given by the function f(x) =15+3600xf ( x ) = 15 + \frac { 3600 } { x }f(x) =15+x3600​ . Find the number of phones that must be produced for the average cost to be $24\$ 24$24 .

Solve. -Five divided by the difference of a number and 9 equals the quotient of 10 and the sum of the number and 11. Find the number.

Solve the equation for the specified variable. - f=f1f2f1+f2\mathrm { f } = \frac { \mathrm { f } _ { 1 } \mathrm { f } _ { 2 } } { \mathrm { f } _ { 1 } + \mathrm { f } _ { 2 } } \quadf=f1​+f2​f1​f2​​ for f2\mathrm { f } _ { 2 }f2​

Solve the equation for the specified variable. - A=h(a+b) 2A = \frac { h ( a + b ) } { 2 } \quadA=2h(a+b) ​ for aaa