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3 years ago

Simplify -

1 answer:

6 0

C. 1/2^-6 x 2^8 (2^8 = 2*2*2*2*2*2*2*2 = 256)
 1/2^-6 x 256 (when ever you do something to the power of a negative you do 1/something to the power of the number.An example is 2^-2 = 1/2^2)
so 1/2^-6 = 1/1/2^6 or 32 (message me if you want me to explain how I got that)
32 x 256
 8,192 <---- answer


d. -(-2)^0 x (-2)^3 x (-2)^-4 divide (-2)^3 (anything ^0 = 1)
 -(1) x -8 x (-2)^-4 divide -8 (-2^-4 = 1/-2^4)
-1 x -8 x 1/16 divide -8 (simplify)
8/16 ÷ -8
 16/8 x -8
 -16 <-- answer

e. I do not know the answers. Sorry

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Step-by-step explanation:

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3 years ago

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3 years ago

What is the solution to the equation -4(2x+3) = 2x+6-(8x+2)?

aivan3 [116]

Answer:

x = -8

Step-by-step explanation:

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Distribute

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-8x-12 = -6x+4

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3 years ago

Hyperbolas

Olenka [21]

let's notice something on this hyperbola, the fraction that is positive, is the fraction with the "y" variable, that simply means that the hyperbola is opening vertically, namely runs over the y-axis or it has a vertical traverse axis, which means, that, the foci will be a certain "c" distance from the center over the y-axis, well, with that mouthful, let's proceed.

$\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ asymptotes\quad y= k\pm \cfrac{a}{b}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\cfrac{(y-3)^2}{1}-\cfrac{(x+2)^2}{4}=1\implies \cfrac{[y-3]^2}{1^2}-\cfrac{[x-(-2)]^2}{2^2}=1~~ \begin{cases} h=-2\\ k=3\\ a=1\\ b=2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ c=\sqrt{a^2+b^2}\implies c=\sqrt{1+4}\implies c=\sqrt{5} \\\\\\ \stackrel{\textit{so then the foci are at}}{(-2~~,~~3\pm \sqrt{5})}\qquad \qquad \qquad \stackrel{\textit{and its vertices are at }}{(-2~~,~~3\pm 1)}\implies \begin{cases} (-2,4)\\ (-2,2) \end{cases}$

now let's check for the asymptotes.

$\bf y=3\pm \cfrac{1}{2}[x-(-2)]\implies y=3\pm \cfrac{1}{2}(x+2) \\\\[-0.35em] ~\dotfill\\\\ y=3+ \cfrac{1}{2}(x+2)\implies y=3+\cfrac{x+2}{2}\implies y=\cfrac{6+x+2}{2} \\\\\\ y=\cfrac{x+8}{2}\implies y=\cfrac{1}{2}x+4 \\\\[-0.35em] ~\dotfill\\\\ y=3- \cfrac{1}{2}(x+2)\implies y=3-\cfrac{(x+2)}{2}\implies y=\cfrac{6-(x+2)}{2} \\\\\\ y=\cfrac{6-x-2}{2}\implies y=\cfrac{-x+4}{2}\implies y=-\cfrac{1}{2}x+2$

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3 years ago

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Korolek [52]

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3 years ago