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1 year ago

9

find the simple interest earned, to the nearest cent, for each principal, interest rate, and time.$750 - 7% - 3 years

1 answer:

Phoenix [80]1 year ago

4 0

We have to apply the simple interest formula:

I = p x r x t

Where :

I = interest

p = principal amount

r= interes rate ( in decimal form, percentage divided by 100)

t= time (years)

Replacing with the values given:

I = 750 x (7/100) x 3

I = 750 x 0.07 x 3

I = $157.7

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Proofs contained within the explanation.

Step-by-step explanation:

These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.

a)

Proof

Base case:

We want to shown the given equation is true for n=1:

The first term on left is 2 so when n=1 the sum of the left is 2.

Now what do we get for the right when n=1:

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Given this assumption we want to show the following:

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Let's start with the left hand side:

$2+5+8+\cdots+(3(k+1)-1)$

$2+5+8+\cdots+(3k-1)+(3(k+1)-1)$

The first k terms we know have a sum of .5k(3k+1) by our assumption.

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Distribute for the second term:

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Combine terms in second term:

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Factor out a half from both terms:

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Distribute for both first and second term in the [ ].

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Combine like terms in the [ ].

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The thing inside the [ ] is called a quadratic expression. It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).

Those numbers would be 3 and 4 since

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$\frac{1}{2}[3k^2+3k+4k+4]$

$\frac{1}{2}[3k(k+1)+4(k+1)]$

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//

b)

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$1+5+5^2+\cdots+5^{k-1}+5^{(k+1)-1}$

We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:

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Apply law of exponents:

$\frac{1}{4}(5^{k+1}-1)$

Therefore the following equation holds for all natural n:

$1+5+5^2+\cdots+5^{n-1}=\frac{1}{4}(5^n-1)$ .

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