Suppose that 18 inches of wire costs 72 cents.

Mathematics

1 answer:

frozen [14]2 years ago

4 0

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Can someone help simplify?: -8x^3y2(2x^4y^4)

Kisachek [45]

So the rule with multiplying exponents of the same base is to add the exponents together. Applying this rule here would look like this:

$-8x^3y^2*2x^4y^4=(-8*2)x^{3+4}y^{2+4}=-16x^7y^6$

In short, your simplified expression is going to be -16x^7y^6.

3 0

3 years ago

You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%

bezimeni [28]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}$

$\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}$