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2 years ago

If Mrs. Hill takes out a mortgage for $80,000 with a 4% interest rate, what will her interest be after 20 years? Help ASAP

Mathematics

1 answer:

Tanzania [10]2 years ago

7 0

Answer:

$6400

Step-by-step explanation:

If Mrs. Hill takes out a mortgage for $80,000 with a 4% interest rate, what will her interest be after 20 years?

First you want to find how much she will get each year.

Turn 4% into a decimal.

Which you have to move the decimal point over to spaces to the left.

4. ----- .4 ----- .04

Now that you have done that you need to multiply,

$80000 × .04

which is $3200.

Now you know she gets $3200 a year.

Now multiply $3200 by 20.

$3200 × 20

which is $6400.

She will get $6400 interest.

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You buy a 33-pound bag of flour for $9 or you can buy a 1- pound bag for $0.39. Compare the per pound cost for the large and sma

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Answer:

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Step-by-step explanation:

9 dollars / 33 lbs = .272727 dollars per lb

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vesna_86 [32]

Using proportions, it is found that the measure of the inscribed angle B is of 30º.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

In a circle, the inscribed angle is half the measure of the outside angle. Hence, in this problem, the measure of angle B is of 50% of 60º, hence:

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2 years ago

Which of the following statements describes a correct method for solving the equation below?

Andrei [34K]

30 + 2w = 2 • (w - 15)

Equation at the end of step 1 :
Step 2 :
Equations which are never true :
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This equation has no solution.
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3 years ago

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Alan has 90 apple tree he has 90 apples and each apple tree how many apples does he have in all?

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3 years ago

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Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

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