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daser333 [38]
3 years ago
13

Someone please help

Mathematics
1 answer:
kkurt [141]3 years ago
7 0
First, make y negative

y=-x-1

From there, you would go down 1 and right 1

3 points you could choose are (0,-1), (-2,1), and (-3,2)
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For the following amount at the given interest rate compounded​ continuously, find​ (a) the future value after 9 ​years, (b) the
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Answer:

(a) The future value after 9 ​years is $7142.49.

(b) The effective rate is r_E=4.759 \:{\%}.

(c) The time to reach ​$13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of P dollars is invested at a rate of interest r compounded continuously for t years, the compound amount is

A=Pe^{rt}

(a) From the information given

P=4700

r=4.65\%=\frac{4.65}{100} =0.0465

t=9 \:years

Applying the above formula we get that

A=4700e^{0.0465\cdot 9}\\A=7142.49

The future value after 9 ​years is $7142.49.

(b) The effective rate is given by

r_E=e^r-1

Therefore,

r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}

(c) To find the time to reach ​$13,000, we must solve the equation

13000=4700e^{0.0465\cdot t}

4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88

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