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agasfer [191]
3 years ago
8

Jenny borrowed $500 for five years at 4 percent interest, compounded annually. What is the total amount she will have paid when

she pays off the loan? total amount = P (1 + i)t $608.33 $729.99 $765.77
Mathematics
1 answer:
joja [24]3 years ago
8 0
P= 500. I = 0.04. t=5. plug those numbers into the formula. So 1.04^5 and multiplying it time 500 and the final answer would be 608.33
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5/9, 2/7, 3/10 are all rational numbers

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Mr. George has a bank balance of -42 dollars at the start of the month. After he deposits 6 dollars, the new balance is (with fu
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3 years ago
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<h2>351.88 m</h2>

Step-by-step explanation:

Given,

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= 2 ( length of semi-circle + length of straight part )

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2 years ago
find the smallest number of terms of the AP "-54,-52.5,-51,-49.5" ....that must be taken for the sum of the terms to be positive
wel

The smallest number of terms of the AP that will make the sum of terms positive is 73.

Since we need to know the number for the sum of terms, we find the sum of terms of the AP

<h3>Sum of terms of an AP</h3>

The sum of terms of an AP is given by S = n/2[2a + (n - 1)d] where

  • n = number of terms,
  • a = first term and
  • d = common difference

Since we have the AP "-54,-52.5,-51,-49.5" ....", the first term, a = -54 and the second term, a₂ = -52.5.

The common difference, d = a₂ - a

= -52.5 - (-54)

= -52.5 + 54

= 1.5

<h3>Number of terms for the Sum of terms to be positive</h3>

Since we require the sum of terms , S to be positive for a given number of terms, n.

So, S ≥ 0

n/2[2a + (n - 1)d] ≥ 0

So, substituting the values of the variables into the equation, we have

n/2[2(-54) + (n - 1) × 1.5] ≥ 0

n/2[-108 + 1.5n - 1.5] ≥ 0

n/2[1.5n - 109.5] ≥ 0

n[1.5n - 109.5] ≥ 0

So, n ≥ 0 or 1.5n - 109.5 ≥ 0

n ≥ 0 or 1.5n ≥ 109.5

n ≥ 0 or n ≥ 109.5/1.5

n ≥ 0 or n ≥ 73

Since n > 0, the minimum value of n is 73.

So, the smallest number of terms of the AP that will make the sum of terms positive is 73.

Learn more about sum of terms of an AP here:

brainly.com/question/24579279

#SPJ1

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