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Juliette [100K]
3 years ago
7

Compute the exact interest on $5,870 at 12% if the money is borrowed from June to December of the same year.

Mathematics
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

The exact interest on $5,870 at 12% is $410.9

Step-by-step explanation:

From the information provided we know that

Principal amount: $5,870

Interest rate: 12% -> 0.12

Time: 7 months (From June to December)

When you know the principal amount, the rate, and the time, the amount of interest can be calculated by using the formula:

I=P\cdot r\cdot t

where P is principal, r is the rate of interest and t is the time in years.

We need to convert the 7 months into 1 year.

7 \>months \cdot \frac{1 year}{12 months} = \frac{7}{12} year

Now we can use the above formula

I=P\cdot r\cdot t=5870\cdot 0.12 \cdot \frac{7}{12} \\I = 0.12\cdot \frac{7\cdot \:5870}{12}\\I = 0.12\cdot \frac{20545}{6}\\I = \frac{2465.4}{6}\\I = 410.9

Therefore the exact interest on $5,870 at 12% is $410.9

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Sam paid $ 124.032 as tax deduction each week.

<h3>What is an Equation ?</h3>

An equation is a mathematical statement formed when an algebraic expression is equated by an equal sign by a constant or algebraic expression.

It is given that

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Let the amount of Tax deduction is represented by $x

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x = 13.6 *912 /100

x = $ 124.032

Therefore Sam paid $ 124.032 as tax deduction each week.

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Step-by-step explanation:

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To find the inverse, we swap the variables y and x, then solve for the new y.

3a. y=\frac{3}{x-1}

Swapping the variables: x=\frac{3}{y-1}
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The domain of this inverse is x ≠ 0.
3b. y=x^2-1

Swapping: x = y^2 - 1
Solving for y: y^2 = x + 1 \\ y = \sqrt{x+1}
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4a. y=\frac{3}{x-1}, y=1+\frac{3}{x}
y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x
y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x

4c. y=\sqrt[3]{\frac{x-7}{3}}, y=3x^3+7
y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x
y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x



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3 years ago
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