Which choice is equivalent to (4^2)(4^3)? A: 4^6 B:4^5 C:64 D:16

2 answers:

6 0

B. 4^5

When multiplying exponents with the same base, you can simply add them together and keep the base the same.
Many people accidentally multiply the exponents, so just be cautious of that :) hope this helps

Taya2010 [7]3 years ago

3 0

Correct me if i'm wrong, but i think it's b, since you add the exponents and keep the 4 the same.

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Tomtit [17]

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5 0

4 years ago

The diameter of a circle has endpoints (-10,5) and (-4,11)

Marta_Voda [28]

well, since we know the endpoints for the diameter, its midpoint will be where the center of the circle is located, so

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -4 -10}{2}~~~ ,~~~ \cfrac{ 11 + 5}{2} \right)\implies \left(\cfrac{-14}{2}~~,~~\cfrac{16}{2} \right)\implies \stackrel{center}{(-7~~,~~8)}$

and if we get the distance between those endpoints, and take half of that, that'd be the radius of it.

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-4 - (-10)]^2 + [11 - 5]^2}\implies d=\sqrt{(-4+10)^2+6^2} \\\\\\ d=\sqrt{6^2+6^2}\implies d=\sqrt{72}~\hfill \stackrel{radius=half~that}{\cfrac{\sqrt{72}}{2}} \\\\[-0.35em] ~\dotfill$

$\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-7}{ h},\stackrel{8}{ k})\qquad \qquad radius=\stackrel{\frac{\sqrt{72}}{2}}{ r} \\\\\\\ [x-(-7)]^2~~ + ~~[y-8]^2~~ = ~~\left( \frac{\sqrt{72}}{2} \right)^2\implies (x+7)^2~~ + ~~(y-8)^2~~ = ~~18$