Question 1

The amount of tread left on a tire varies inversely as the number of miles the tire has traveled. A tire that has traveled  40,000 mi has  $\frac { 5 } { 8 }$  in. of tread left. How much tread will be left on a tire that has traveled  19,000 mi ?

Accepted Answer

The relationship between tread left (T) and miles traveled (M) is inversely proportional, so $T = k/M$, where k is a constant. For a tire with 40,000 miles and $\frac{5}{8}$ inch of tread, $k = T \times M = \frac{5}{8} \times 40,000$. For 19,000 miles, using the same k, solve for T to find the tread left. The correct calculation leads to $\frac{25}{19}$ inches of tread left.

Question 2

Solve the problem. Round your answer to two decimal places.
-The weight W of an object on the Moon varies directly as the weight E on earth. A person who weighs 136 lb on earth weighs 27.2 lb on the Moon. How much would a 124-lb person weigh on the Moon?

Accepted Answer

The weight on the Moon can be found using the ratio 27.2/136, which is the weight on the Moon to the weight on Earth for the given person. For a 124-lb person, the weight on the Moon would be 124 * (27.2/136) = 24.8 lb.

Question 3

Solve the problem. Round your answer to two decimal places.
-The distance to the horizon varies directly as the square root of the height above ground level of the observer. If a person can see 6 miles from a height of 25 feet, how far can a person see from a height of 64 feet?

Accepted Answer

The answer of Solve the problem. Round your answer to...

Question 4

Solve the problem. Round your answer to two decimal places.
-The distance an object falls when dropped from a tower varies directly as the square of the time it falls. If the object falls 144 feet in 3 seconds, how far will it fall in 15 seconds?

Accepted Answer

The distance $d$ varies directly as the square of the time $t$, so $d = kt^2$ for some constant $k$. Given $d = 144$ feet when $t = 3$ seconds, we find $k = 144 / 3^2 = 144 / 9 = 16$. For $t = 15$ seconds, $d = 16(15^2) = 16(225) = 3600$ feet.

Question 5

$$\text { If } \mathrm { f } \text { varies jointly as } \mathrm { q } ^ { 2 } \text { and } h \text {, and } \mathrm { f } = 36 \text { when } \mathrm { q } = 2 \text { and } h = 3 \text {, find } \mathrm { f } \text { when } \mathrm { q } = 4 \text { and } \mathrm { h } = 6 \text {. }$$

Accepted Answer

The formula for joint variation is $f = kq^2h$, where $k$ is the constant of variation. Given $f = 36$, $q = 2$, and $h = 3$, we can find $k$ by substituting these values: $36 = k(2)^2(3)$, which simplifies to $36 = 12k$, so $k = 3$. To find $f$ when $q = 4$ and $h = 6$, substitute these values and $k = 3$ into the formula: $f = 3(4)^2(6) = 3(16)(6) = 288$.

Question 6

$$x \text { varies inversely as } v \text {, and } x = 45 \text { when } v = 5 \text {. Find } x \text { when } v = 25 \text {. }$$

Accepted Answer

The relationship between $x$ and $v$ is given by $x = k/v$, where $k$ is a constant. When $x = 45$ and $v = 5$, we can find $k = 45 \times 5 = 225$. To find $x$ when $v = 25$, we use $x = 225 / 25 = 9$.

Question 7

The current I in an electrical conductor varies inversely as the resistance R of the conductor. The current is 5 amperes when the resistance is 888 ohms. What is the current when the resistance is 550 ohms? Round to the
Nearest tenth.

Accepted Answer

The relationship between current (I) and resistance (R) is given by I = k/R, where k is a constant. For I = 5 amperes and R = 888 ohms, we find k = 5 * 888. To find the current when R = 550 ohms, we use the same k and solve for I, resulting in I = k / 550, which gives approximately 8.1 amperes.

Question 8

If f varies jointly as $\mathrm { q } ^ { 2 }$ and  h, and f=144  when q=4  and h=3 , find  q when  f=72  and h=6 .

Accepted Answer

The joint variation equation is $ f = k \cdot q^2 \cdot h $. Given $ f = 144 $, $ q = 4 $, and $ h = 3 $, we find $ k = \frac{144}{4^2 \cdot 3} = 3 $. For $ f = 72 $ and $ h = 6 $, solve $ 72 = 3 \cdot q^2 \cdot 6 $ to find $ q^2 = 4 $, so $ q = 2 $.

Question 9

The gravitational attraction A between two masses varies inversely as the square of the distance between them. The force of attraction is 4 lb when the masses are 3 ft apart, what is the attraction when the masses are 6 ft
Apart?

Accepted Answer

The gravitational force between two masses is given by the formula $F = G \frac{m_1 m_2}{r^2}$, where $F$ is the force of attraction, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the centers of the two masses. Since the force varies inversely as the square of the distance, when the distance is doubled (from 3 ft to 6 ft), the force is reduced by a factor of $2^2 = 4$. Therefore, the new force of attraction is $4 lb / 4 = 1 lb$.

Question 10

Round to the nearest tenth unless indicated otherwise.
-The volume $\mathrm { V }$ of a given mass of gas varies directly as the temperature $\mathrm { T }$ and inversely as the pressure $\mathrm { P }$. If $\mathrm { V } =$ $180.0$ in $^ { 3 }$ when $\mathrm { T } = 300 ^ { \circ }$ and $\mathrm { P } = 20 \mathrm { lb } / \mathrm { in } ^ { 2 }$, what is the volume when $\mathrm { T } = 210 ^ { \circ }$ and $\mathrm { P } = 15 \mathrm { lb } / \mathrm { in } ^ { 2 }$ ?

Accepted Answer

The volume of a gas varies directly with temperature and inversely with pressure, so $V = k \frac{T}{P}$ for some constant $k$. Using the given values to solve for $k$ and then applying it to find the new volume gives the correct answer.

Question 11

The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth. If a person weighs 180 pounds on the surface of the earth and the radius
Of the earth is 3900 miles, what will the person weigh if he or she is 150 miles above the earth's surface? Round
Your answer to the nearest hundredth of a pound.

Accepted Answer

The weight varies inversely as the square of the distance from the center of the Earth. The formula for weight at a distance $r$ from the center is $W = \frac{k}{r^2}$, where $k$ is a constant. At the surface, $r = 3900$ miles, so $180 = \frac{k}{3900^2}$. To find the weight 150 miles above the surface, $r = 4050$ miles. Using the inverse square law, the weight at 4050 miles from the center is calculated and rounded to the nearest hundredth, resulting in 166.91 pounds.

Question 12

The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station. Suppose the the intensity is 8000 units at a distance of 2 miles. What will the intensity be at a distance of
5 miles? Round your answer to the nearest unit.

Accepted Answer

The answer of The intensity of a radio signal from...

Question 13

The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a screen 5 ft from a light is 2 foot candles, find the intensity on a screen 20 ft from the light.

Accepted Answer

The intensity of light varies inversely as the square of the distance, so $I_1D_1^2 = I_2D_2^2$. Given $I_1 = 2$ foot-candles at $D_1 = 5$ ft, and $D_2 = 20$ ft, we find $I_2$ by substituting: $2 \times 5^2 = I_2 \times 20^2$, which simplifies to $I_2 = \frac{1}{8}$ foot-candle.

Question 14

Solve the problem. Round your answer to two decimal places.
-The area of a circle varies directly as the square of the radius of the circle. If a circle with a radius of 5 inches has an area of 78.5 square inches, what is the area of a circle with a radius of 18 inches?

Accepted Answer

The answer of Solve the problem. Round your answer to...

Question 15

Solve the problem. Round your answer to two decimal places.
-The distance it takes to stop a car varies directly as the square of the speed of the car. If it takes 112 feet for a car traveling at 40 miles per hour to stop, what distance is required for a speed of 45 miles per hour?

Accepted Answer

The stopping distance varies directly with the square of the speed, so we use the formula $ d = k \cdot s^2 $. First, find $ k $ using the given data: $ 112 = k \cdot 40^2 $. Solving for $ k $, we get $ k = \frac{112}{1600} = 0.07 $. For 45 mph, $ d = 0.07 \cdot 45^2 = 141.75 $ feet.

Question 16

The time it takes to complete a certain job varies inversely to the number of people working on that job. If it takes 28 hours for 11 carpenters to frame a house, then how long will it take 49 carpenters to do the same job?

Accepted Answer

The time it takes to complete a job varies inversely with the number of people, so $T = k/N$, where $T$ is time, $N$ is the number of people, and $k$ is a constant. Given $T = 28$ hours for $N = 11$ carpenters, we find $k = 28 \times 11$. For $N = 49$ carpenters, $T = k/49 = (28 \times 11)/49 = 6.3$ hours.

Question 17

The weight of a body above the surface of the earth is inversely proportional to the square of its distance from the center of the earth. What is the effect on the weight when the distance is multiplied by 2?

Accepted Answer

Since weight is inversely proportional to the square of the distance, doubling the distance results in the weight being divided by $2^2 = 4$.

Question 18

The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 254 vibrations per second and a wavelength of 19.2 ft. Find the wavelength of another tone that has a pitch of 358 vibrations per
Second. Round to the nearest tenth.

Accepted Answer

The relationship between pitch (P) and wavelength (W) is inversely proportional, meaning $P = k/W$, where $k$ is a constant. For the first tone, $254 = k/19.2$, so $k = 254 \times 19.2$. For the second tone with a pitch of 358 vibrations per second, $358 = k/W$. Substituting the value of $k$ from the first equation, we find the wavelength $W$ for the second tone. Solving for $W$ gives $W \approx 13.6$ ft.

Question 19

The weight that a horizontal beam can support varies inversely as the length of the beam. Suppose that a 5-m beam can support 480 kg. How many kilograms can a 2-m beam support?

Accepted Answer

The weight a beam can support is inversely proportional to its length, meaning if the length is halved, the weight it can support is doubled. Using the formula $W_1 \times L_1 = W_2 \times L_2$, where $W_1 = 480$ kg and $L_1 = 5$ m (for the first beam), and $L_2 = 2$ m (for the second beam), we solve for $W_2$, the weight the 2-m beam can support. Substituting the known values, $480 \times 5 = W_2 \times 2$. Solving for $W_2$ gives $W_2 = \frac{480 \times 5}{2} = 1200$ kg.

Question 20

$$\text { If } x \text { varies inversely as } y ^ { 2 } \text {, and } x = 2 \text { when } y = 20 \text {, find } x \text { when } y = 5 \text {. }$$

Accepted Answer

The answer of \[\text { If } x \text {...

Quiz 4: Polynomials and Rational Functions

The amount of tread left on a tire varies inversely as the number of miles the tire has traveled. A tire that has traveled 40,000 mi has 58\frac { 5 } { 8 }85​ in. of tread left. How much tread will be left on a tire that has traveled 19,000 mi ?

Solve the problem. Round your answer to two decimal places. -The weight W of an object on the Moon varies directly as the weight E on earth. A person who weighs 136 lb on earth weighs 27.2 lb on the Moon. How much would a 124-lb person weigh on the Moon?

Solve the problem. Round your answer to two decimal places. -The distance to the horizon varies directly as the square root of the height above ground level of the observer. If a person can see 6 miles from a height of 25 feet, how far can a person see from a height of 64 feet?

Solve the problem. Round your answer to two decimal places. -The distance an object falls when dropped from a tower varies directly as the square of the time it falls. If the object falls 144 feet in 3 seconds, how far will it fall in 15 seconds?

x varies inversely as v, and x=45 when v=5. Find x when v=25. x \text { varies inversely as } v \text {, and } x = 45 \text { when } v = 5 \text {. Find } x \text { when } v = 25 \text {. }x varies inversely as v, and x=45 when v=5. Find x when v=25.

The current I in an electrical conductor varies inversely as the resistance R of the conductor. The current is 5 amperes when the resistance is 888 ohms. What is the current when the resistance is 550 ohms? Round to the Nearest tenth.

If f varies jointly as q2\mathrm { q } ^ { 2 }q2 and h, and f=144 when q=4 and h=3 , find q when f=72 and h=6 .

The gravitational attraction A between two masses varies inversely as the square of the distance between them. The force of attraction is 4 lb when the masses are 3 ft apart, what is the attraction when the masses are 6 ft Apart?

The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station. Suppose the the intensity is 8000 units at a distance of 2 miles. What will the intensity be at a distance of 5 miles? Round your answer to the nearest unit.

The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a screen 5 ft from a light is 2 foot candles, find the intensity on a screen 20 ft from the light.

Solve the problem. Round your answer to two decimal places. -The area of a circle varies directly as the square of the radius of the circle. If a circle with a radius of 5 inches has an area of 78.5 square inches, what is the area of a circle with a radius of 18 inches?

Solve the problem. Round your answer to two decimal places. -The distance it takes to stop a car varies directly as the square of the speed of the car. If it takes 112 feet for a car traveling at 40 miles per hour to stop, what distance is required for a speed of 45 miles per hour?

The time it takes to complete a certain job varies inversely to the number of people working on that job. If it takes 28 hours for 11 carpenters to frame a house, then how long will it take 49 carpenters to do the same job?

The weight of a body above the surface of the earth is inversely proportional to the square of its distance from the center of the earth. What is the effect on the weight when the distance is multiplied by 2?

The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 254 vibrations per second and a wavelength of 19.2 ft. Find the wavelength of another tone that has a pitch of 358 vibrations per Second. Round to the nearest tenth.

The weight that a horizontal beam can support varies inversely as the length of the beam. Suppose that a 5-m beam can support 480 kg. How many kilograms can a 2-m beam support?

If x varies inversely as y2, and x=2 when y=20, find x when y=5. \text { If } x \text { varies inversely as } y ^ { 2 } \text {, and } x = 2 \text { when } y = 20 \text {, find } x \text { when } y = 5 \text {. } If x varies inversely as y2, and x=2 when y=20, find x when y=5.

The amount of tread left on a tire varies inversely as the number of miles the tire has traveled. A tire that has traveled 40,000 mi has $\frac { 5 } { 8 }$ in. of tread left. How much tread will be left on a tire that has traveled 19,000 mi ?

$x \text { varies inversely as } v \text {, and } x = 45 \text { when } v = 5 \text {. Find } x \text { when } v = 25 \text {. }$

If f varies jointly as $\mathrm { q } ^ { 2 }$ and h, and f=144 when q=4 and h=3 , find q when f=72 and h=6 .

$\text { If } x \text { varies inversely as } y ^ { 2 } \text {, and } x = 2 \text { when } y = 20 \text {, find } x \text { when } y = 5 \text {. }$