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Digiron [165]
3 years ago
12

Need help -5(1+2k)-8(-4+5k)

Mathematics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

27 - 50k

Simplify

1. Distribute

-5 ( 1 + 2k ) - 8 ( -4 + 5k )

-5 - 10k - 8 ( -4 + 5k )

2. Distribute

-5 - 10k - 8 ( -4 + 5k )

-5 - 10k + 32 - 40k

3. Add the numbers

-5 - 10k + 32 - 40k

27 - 10k - 40k

4. Add the same term to both sides of the equation

27 - 10k - 40k

27 - 50k

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What is the value of x?<br><br><br><br> Enter your answer in the box.<br> x =<br> ​​
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Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How
Scorpion4ik [409]

let's check how much is it after 2 years firstly.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275


Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

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3 years ago
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