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3 years ago

Zeke spent 12 minutes on the phone while routing 4 phone calls. In all, how many phone

Mathematics

1 answer:

Keith_Richards [23]3 years ago

3 0

5 because
12/4 = 15/x
X=5

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Write 250 million in standard form.

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250,000,000 or 250000000

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K/π = 6.7 (Solve For k)

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A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming tha

White raven [17]

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

$\frac{dB}{dt} = rB-k$

$\Rightarrow \frac{dB}{dt} - rB=-k$ .......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor $=e^{\int p(t) dt$

$=e^{\int (-r)dt$

$=e^{-rt}$

Multiplying the integrating factor the both sides of equation (1)

$e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}$

$\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt$

Integrating both sides

$\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt$

$\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C$ [ where C arbitrary constant]

$\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}$

Initial condition B=7864 when t =0

$\therefore 7864= \frac{k}{r} - Ce^0$

$\Rightarrow C= \frac{k}{r} -7864$

Then the general solution is

$B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}$

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

$B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}$

$\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0$

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

=$1573.17

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4 years ago

I’m stuck please help me thanks ! The solids are similar. Find the volume of prism B.

lesantik [10]

Answer:

I think it's 1,728 millimeters ^3

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