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Len [333]
3 years ago
6

The constant of proportionality for this relationship is

Mathematics
2 answers:
Alex73 [517]3 years ago
6 0

Answer:

.4

Step-by-step explanation:

Vesnalui [34]3 years ago
4 0
The answer for this question is 4
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The one in the top left corner is the answer

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1. True or False: Please answer the questions in the boxes provided.
vivado [14]

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trufe

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Fifteen years ago, you deposited $12,500 into an investment fund. Five years ago, you added an
True [87]

Answer:

a) 8.16%

b) $65,762.50

c) $39,701.07

Step-by-step explanation:

Given:

15 years ago, Initial investment = $12500

5 years ago, Investment = $20000

Nominal interest = 8% semi annually for first 10 years

Interest2= 6.5% compunded annually for last five years

a) for the effective annual interest rate (EAR) in the first 10 years, let's use the formula:

[1+(nominal interest rate/number of compounding periods)]^ number of compounding periods-1

EAR = [1 + (\frac{0.08}{2})]^2 - 1

= (1 + 0.04)^2 - 1

= (1.04)^2 - 1

= 1.0816 - 1

= 0.0816

= 8.16%

The effective annual interesting rate, EAR = 8.16 %

b) To find the amount in my account today.

Let's first find the amount for $12500 for 10 years compounded semi annually

= 12,500 +( 12,500 * 8.160% * 10)

= $ 22,700

Let's also find the amount for $32,500($12,500+$20,000) for 5 Years compoundeed annually

$32,500 + ($32,500 * 6.5% *5)

= $ 43,062.50

Money in account today will be:

$22,700 + $43,062.50

= $65,762.50

c) Let's the amount I should have invested to be X

For first 10 years at 8.160%, we have:

Interest Amount = ( X * 8.160% * 10 ) = 0.8160 X

For next 5 years at 6.5%, we have:

Interest Amount = (X * 6.5% * 5) = 0.325 X

Therefore the total money at the end of 15 Years = 85000

0.8160X + 0.3250X + X = $85,000

= 2.141X = $85,000

X = 85,000/2.141

X = 39,701.074 ≈ $39,700

If I wish to have $85,000 now, I should have invested $39,700 15 years ago

7 0
3 years ago
Find f'(x) for f(x) = cos^2(3x^3).
garik1379 [7]

f(x) = \cos^2(3x^3)\\f'(x) = \frac{d}{dx}[\cos^2(3x^3)]\\= 2\cos(3x^3) \cdot \frac{d}{dx}[cos(3x^3)]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot \frac{d}{dx}[3x^3]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot 9x^2\\= -18x^2\cos(3x^3)\sin(3x^3)

6 0
3 years ago
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