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4 years ago

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Ryan is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. Compa

ny A charges an initial fee of $72.50 and an additional 40 cents for every mile driven. Company B charges an initial fee of $25 and an additional 90 cents for every mile driven. For what mileages will Company A charge no more than Company B? Write your answer as an inequality, using m for the number of miles driven.

1 answer:

son4ous [18]4 years ago

3 0

m≥95

Explanation:

Since, we have two option given.

forming the equation for Company A

As company A pay fixed (intercept) of $72.5. Moreover, with every one mile driven company A pays $0.4 (slope)

Using the equation

y=mx+c

where m is slope, and c is intercept.

In the case of company A. slope is $0.4, and intercept is $72.5.

charges= 0.4m+72.5

Forming the equation for company B

charges=0.9m+25

Now, as per the requirement of question we must find the value of m, where Company A will charge no more than company B

<em>That means,</em>

<em>we have to find the value of m where charges from equation of company A should be less than or equal to charges from equation of company B</em>

<em>in other words,</em>

<em>0.4m+72.5 ≤ </em> 0.9m+25

solving for m,

Step 1

72.5-25 ≤ 0.9m-0.4m

Step 2

47.5≤ 0.5m

Step 3

47.5/0.5≤ m

Step 4

95≤ m

in other words, m≥95

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A body in the solar system has a period of 10,759.22 days and a perihelion speed of 10.18 km/s. a. Calculate the aphelion radius

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Explanation:

From the information given, by applying Kepler's 3rd law,

$T^2 \alpha a^3$

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T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 $\times 10^8 \ km$

Therefore, $T^2 = ca^3$

$c= \dfrac{365^2}{(1.496 \times 10^8)^3}$

c = 3.9791 $\times 10^{20} \ day^2/km^3$

However, if the body in the solar system has a period of 10.759.22 days, then, a =?

∴

$T^2 = ca^3$

$a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}$

$a^3 = 2.9092 \times 10^{27}$

$a= \sqrt[3]{2.9092 \times 10^{27}}$

a = 1.4275 $\times 10^9 \ km$

However, the velocity for a perihelion = 10.18 km/s

Using the formula

$v = \sqrt{GM ( \dfrac{2}{r}-\dfrac{1}{a})}$ to calculate the radius, we have:

G = $6.674 \times 10^{-11}$

M = $1.989\times 10^{30} \ kg$

r = perihelion

$v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})$

$(10.18 \times 10^3) ^2= 6.674 \times 10^{-11} \times 1.989 \times 10^{30} ( \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}})$

$7.8068 \times 10^{-13}= \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}}$

$\dfrac{2}{r} = 1.4824 \times 10^{-12}$

$r = \dfrac{2}{1.4824 \times 10^{-12}}$

$r = 1.349 \times 10^{12}$

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

∴

$1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)$

$1.349 \times 10^{12} - 1.425 \times 10^{12}= - 1.425 \times 10^{12} (e)$

$-7.6\times 10^{10}= - 1.425 \times 10^{12} (e)$

$\dfrac{-7.6\times 10^{10}}{- 1.425 \times 10^{12}}= (e)$

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$r = 1.425 \times 10^{12} ( 1 + 0.0533)$

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