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2 years ago

You deposit $500 into an account that pays 4.25% annual interest compounded continuously. What is the balance after 10 years?

1 answer:

6 0

$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$500\\ r=rate\to 4.25\%\to \frac{4.25}{100}\dotfill &0.0425\\ t=years\dotfill &10 \end{cases} \\\\\\ A=500e^{0.0425\cdot 10}\implies A=500e^{0.425}\implies A\approx 764.80$

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A bank loaned out $21,500, part of it at the rate of 15% annual interest, and the rest at 9% annual interest. The total interest

Nuetrik [128]

Answer:

So the portion of the loan lent at 4% is 21000(0.5955) = $12,505.5

And the portion of the loan lent at 15% is 21000(0.4045) = $8494.5

Step-by-step explanation:

This problem can be solved in several ways. The way I like is as follows.

Ask yourself at what interest would $ 21,000 yield and interest of $ 2,215 in one year ? The equation to solve is:

21000(P) = 2215

P = 0.1055 or 10.55 %

This number is a weighed average of the 4% yield and the 15% yield so the next equation to solve is

10.55 = 4(1 – X) + 15X

Where X is the proportion of the loan given at the indicated interest. So

10.55 = 4 – 4X + 15X

6.55 = 11X

X = 0.5955

1-X = 0.4045

So the portion of the loan lent at 4% is 21000(0.5955) = $12,505.5

And the portion of the loan lent at 15% is 21000(0.4045) = $8494.5

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3 years ago

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AfilCa [17]

Answer:

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Step-by-step explanation:

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natta225 [31]

Answer:

(a) at t=2, the particle is moving toward the origin

(b) a(t) = 70(8 -9t^2)/(3t^2 +8)^3; a(2) = -0.245

Step-by-step explanation:

(a) The function x(t) is negative for -3 < t < 3, so at t = 2, the particle is to the left of the origin.

The velocity of the particle is given by the derivative of the position function:

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(b) The acceleration is the derivative of the velocity, so is ...

$a(t)=\dfrac{70(3t^2+8)^2-(70t)(12t)(3t^2+8)}{(3t^2+8)^4}=\dfrac{70(3t^2+8-12t^2)}{(3t^2+8)^3}\\\\\boxed{a(t)=\dfrac{70(8-9t^2)}{(3t^2+8)^3}}$

Then at t=2, the acceleration is ...

a(2) = 70(8 -9·4)/(3·4+8)^3 = 70(-28)/8000

a(2) = -0.245

(c) As t gets large, the value of x(t) approaches t^2/(3t^2) = 1/3.

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