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Harlamova29_29 [7]

3 years ago

15

Ace Cars says they will pay Jim a 10% commission on the sale price of each car. Budget Cars says they will pay him $500 per week

plus a 7.5% commission on the sale price of each car. Jim can sell one car every week. What price would the car have to sell for in order for these two job opportunities to pay Jim the same amount?

1 answer:

Arisa [49]3 years ago

8 0

Answer:

$20,000

Step-by-step explanation:

You can quickly figure this by recognizing that the difference in commission must be the same as the base salary:

2.5%(car price) = $500

car price = $500/0.025 = $20,000

__

If you prefer, you can get there with equations.

At Job 1, the pay is ...

y = 10%×(car price)

At Job 2, the pay is ...

y = 7.5%×(car price)

Setting the equal (the two opportunities pay the same), we have ...

0.10(car price) = 500 +0.075(car price)

0.025(car price) = 500 . . . . . . . . subtract 0.075(car price)

car price = 500/0.025 = 20,000 . . . . divide by 0.025

Jim would have to sell a $20,000 car each week for the jobs to pay the same.

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1.64 as mixed number in simplest form

Anettt [7]

Answer:

$\large\boxed{1.64=1\dfrac{16}{25}}$

Step-by-step explanation:

$1.64=1+0.\underbrace{64}_2=1+\dfrac{64}{1\underbrace{00}_2}=1\dfrac{64}{100}=1\dfrac{64:4}{100:4}=1\dfrac{16}{25}$

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3 years ago

Solve the equation 6(3x + 4) = 4x − 4.

emmasim [6.3K]

Answer:

x = - 2

Step-by-step explanation:

Given

6(3x + 4) = 4x - 4 ← distribute parenthesis on left side

18x + 24 = 4x - 4 ( subtract 4x from both sides )

14x + 24 = - 4 ( subtract 24 from both sides )

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7 0

3 years ago

Read 2 more answers

Determine the equation of the circle graphed below.

sergij07 [2.7K]

Answer:

Equation = (x - 6 )² + ( y + 3 )² = 9

Step-by-step explanation:

The circle passes through ( 6, 0) and ( 6 , -6)

They are the coordinates of the diameter.

Using this we can find the centre of the circle.

<u>Find the centre of the circle.</u>

Centre of the circle is the mid- point of (6, 0) and ( 6, -6)

$Centre = (\frac{x_1+x_2}{2} , \frac{y_1 + y_2}{2})$

$=(\frac{6 + 6}{2}, \frac{0 + (-6)}{2})\\\\=(6, -3)$

<u>Find the radius of the circle.</u>

$Radius = \frac{Diameter }{2}$

Diameter is the distance between the points (6 , 0) and ( 6, - 6)

$Diameter = \sqrt{(x_2 - x_1)^2+ (y_2 - y_1)^2\\}$

$=\sqrt{(6-6)^2 + (-6 -0)^2}\\\\=\sqrt{0 + 36} \\\\= 6$

Therefore,

$Radius ,r = \frac{6}{2} = 3$

<u>Standard equation of a circle:</u>

<u></u> $(x - a)^2 + (y - b)^2 = r^2 \ where \ (a , b) \ is \ the\ centre \ coordinates.$ <u></u>

Therefore , equation of the circle ;

$(x - 6)^2 + (y + 3)^2 = 3^2\\\\(x -6)^2 + (y + 3)^2 = 9$

6 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

Plsss help ASAP !! giving the brainlest answer to whoever can help me out?? plsss

timofeeve [1]

Answer:

The 2nd one

Step-by-step explanation:

6 0

3 years ago