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3 years ago

I only need number one. Which variable should be the dependent variable?

Mathematics

1 answer:

Natali [406]3 years ago

5 0

The dependent variable is y, that represents the years

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2 years ago

An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click

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Answer:

A. $\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

B. $\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

Step-by-step explanation:

Given that:

An investment of Amount = $8000

earns at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A) Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

$A = Pe^{rt}$

where;

$A = (8000) \ e ^{0.7t}$

The instantaneous rate of change = $\dfrac{dA}{dt}$

$\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )$

$= 8000 \dfrac{d}{dt}e^{0.07 \ t}$

$\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}$

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 2 years; the instantaneous rate of change is:

$\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}$

$\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

$12000 = (8000)e^{0.07 \ t}$

$\dfrac{12000 }{8000}= e^{0.07 \ t}$

$1.5= e^{0.07 \ t}$

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 5.79

$\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}$

$\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

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