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3 years ago

An agent makes 16% commission on an athlete's signing bonus. If the bonus is $26,000

1 answer:

5 0

Multiply the percentage of commission by the amount of the bonus.

16% = 0.16

26,000 * 0.16 = 4,160

The agent earned $4,160.

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Diego wanted to find the 7% sales tax on purchases that total $15.99. On his calculator the display showed $1.1193 as the tax. H

cricket20 [7]

If I understand this question correctly, Diego will pay $1.12 because the 9 rounds the 1 up to a 2.

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3 years ago

Find the median of the following: 302, 220, 221, and 208. if needed, round your answer to the nearest tenth

MissTica

The Median for this is 220.5

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3 years ago

Read 2 more answers

Write the equation of the quadratic function with roots 0 and 2 and a vertex at (1,5).

lisov135 [29]

Answer:

Step-by-step explanation:

y=a(x-0)(x-2)

=a(x^2-2x)

=a(x^2-2x+1-1)

or y=a(x-1)^2-a

(1,5) lies on it.

5=a(1-1)^2-a

5=0-a

or a=-5

so y=-5(x-1)^2+5

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3 years ago

Factor the Polynomial by solving x^2 + bx + c = 0.<br><br> m^2 - mv - 56v^2

marissa [1.9K]

Factor the Polynomial by solving x^2 + bx + c = 0.

m^2 - mv - 56v^2

Factor by grouping.

(m−8v)(m+7v).

x^2 + bx + c = 0

$x=\frac{-a+\sqrt{a^{2-4c} } }{2} , x=\frac{-a-\sqrt{a^{2-4c} } }{2}$

7 0

3 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

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4 years ago