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Nutka1998 [239]
3 years ago
10

Place the indicated product in the proper location on the grid. (2a ^2b ) ^4

Mathematics
2 answers:
il63 [147K]3 years ago
7 0
<span>(2a ^2b ) ^4
= (2)^4 * (a^2b)^4
= 16 * a^(2b * 4)
= 16a^(8b)

</span>
Anastasy [175]3 years ago
3 0

Answer:

Step-by-step explanation:

Using the two rules: (ab)^x = (a^x)(b^x) and (a^x)^y = a^(xy)

(2a ^2b ) ^4

= (2)^4 x (a^2b)^4

= 16 x a^(2b x 4)

= 16a^(8b)

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Answer:

Part A)

About $3767.34.

Part B)

About $3692.47.

Step-by-step explanation:

Part A)

Recall that compound interest is given by the formula:
\displaystyle A = P\left(1+\frac{r}{n}\right)^{nt}

Where <em>A</em> is the final amount, <em>P</em> is the initial amount, <em>r</em> is the interest rate, <em>n</em> is the number of times compounded per year, and <em>t</em> is the number of years.

To obtain $4000 after two years, let <em>A</em> = 4000 and<em> t</em> = 2.

Because the account pays 3% interest compounded monthly, <em>r</em> = 0.03 and <em>n</em> = 12.

Substitute and solve for <em>P: </em>

<em />\displaystyle \begin{aligned} (4000) & = P\left(1+\frac{(0.03)}{(12)}\right)^{(12)(2)} \\ \\ P & = \frac{4000}{\left(1+\dfrac{(0.03)}{(12)}\right)^{(12)(2)}} \\ \\ & \approx \$3767.34\end{aligned}

In concluion, about $3767.34 should be deposited.

Part B)

Recall the formula for continuous compound:

\displaystyle A = Pe^{rt}

Where <em>e</em> is Euler's number.

Hence, let <em>A</em> = 4000, <em>r</em> = 0.04 and <em>t</em> = 2. Substitute and solve for <em>P: </em>

<em />\displaystyle \begin{aligned}(4000) & = Pe^{(0.04)(2)} \\ \\ P & = \frac{4000}{e^{(0.02)(4)}} \\ \\ & \approx \$3692.47 \end{aligned}

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Hello!

Use the distributive property and multiply like terms.

<h3>Simplify</h3>
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