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kykrilka [37]
2 years ago
6

Tom has a large photo he wants to shrink to wallet-sized. It's width is 20 centimeters and it's len is 30 centimeters. If he wan

ts the width to be 5 centimeters what should the length be?
Mathematics
1 answer:
dsp732 years ago
5 0
We can use ratios, in other words, if one part of the photo is changed x times, how many times does the other part change?

We can set up the proportion as :
width 1 : length 1 = width 2 : length 2

20 : 30 = 5 : x

We do inverse multiplication, or multiply the outer with outer and inner with inner (W1 * L2 and L1 * W2)

20*x = 30*5
20x = 150
Divide all by 10
2x = 15
Divide all by 2 to isolate x
x = 15/2
x = 7.5cm
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Answer:

a) The present value is 688.64 $

b) The accumulated amount is 1532.60 $

Step-by-step explanation:

<u>a)</u><u> The preset value equation is given by this formula:</u>

P=\int^{T}_{0}f(t)e^{-rt}dt

where:

  • T is the period in years (T = 10 years)
  • r is the annual interest rate (r=0.08)

So we have:

P=\int^{T}_{0}(0.01t+100)e^{-rt}dt

Now we just need to solve this integral.

P=\int^{T}_{0}0.01te^{-rt}dt+\int^{T}_{0}100e^{-rt}dt

P=e^{-0.08t}(-1.56-0.13t)|^{10}_{0}+1250e^{-0.08t}|^{10}_{0}

P=0.30+688.34=688.64 $

The present value is 688.64 $

<u>b)</u><u> The accumulated amount of money flow formula is:</u>

A=e^{r\tau}\int^{T}_{0}f(t)e^{-rt}dt

We have the same equation but whit a term that depends of τ, in our case it is 10.

So we have:

A=e^{r\tau}\int^{T}_{0}(0.01t+100)e^{-rt}dt=e^{0.08\cdot 10}P

A=e^{0.08\cdot 10}688.64=1532.60 $

The accumulated amount is 1532.60 $

Have a nice day!

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