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OleMash [197]
4 years ago
15

1.25=0.75+r what is R?

Mathematics
1 answer:
r-ruslan [8.4K]4 years ago
6 0

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

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Write as a single power, then<br> evaluate.<br> (3^3)^4
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<h3>531,441</h3>

Step-by-step explanation:

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2/8 greater or less than 1/2
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a parallelogram has one angle that measures 40. what are the measures of the other three angles in the parallelogram?
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3 years ago
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Suppose yoko borrows $3500 at an interest rate of 4% compounded each year Find the amount owed at the end of 1 year
olchik [2.2K]

Amount owed at the end of 1 year is 3640

<h3><u>Solution:</u></h3>

Given that yoko borrows $3500.

Rate of interest charged is 4% compounded each year

Need to determine amount owed at the end of 1 year.

In our case :

Borrowed Amount that is principal P = $3500

Rate of interest r = 4%

Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1

<em><u>Formula for Amount of compounded yearly is as follows:</u></em>

A=p\left(1+\frac{r}{100}\right)^{n}

Where "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula we get

\mathrm{A}=3500\left(1+\frac{4}{100}\right)^{1}

\begin{array}{l}{A=\frac{3500 \times 104}{100}} \\\\ {A=35 \times 104} \\\\ {A=3640}\end{array}

Hence amount owed at the end of 1 year is 3640

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