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FinnZ [79.3K]
4 years ago
7

Simplify -5(20 + 3). 85 - 115 -97 -103

Mathematics
1 answer:
Alexxandr [17]4 years ago
4 0

Answer:

-115

Step-by-step explanation:

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Step-by-step explanation:

We have been given that it equals 19 when added to 3 and multiplied by 4 so follow these steps:

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This is used to solve for x

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Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.
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\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}

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