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larisa86 [58]
3 years ago
7

identify an equation in point slope form for the line perpendicular to the y=-1/2x+11 that passes through (4,-8). a. y+8=1/2(x-4

) b. y-4=2(x+8) c. y-8=1/2(x+4) d. y+8=2(x-4)
Mathematics
1 answer:
Solnce55 [7]3 years ago
3 0

Answer:

d. y+8=2(x-4)

Step-by-step explanation:

There are 2 important parts to this question. First, understanding which slopes are perpendicular. The negative reciprocal of a number will be perpendicular to it. So, since the original slope is -1/2 the new slope should be 2.

Then, remember what the point-slope formula is. The point-slope formula is: y-y_{2}=m(x-x_{2}). So if you plug in the point and slope the new equation looks like, y--8=2(x-4). Then, simplify for the final answer of y+8=2(x-4).

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3 years ago
P³ = 1/8 please help me if anybody can .
ivanzaharov [21]

Answer:

p= \dfrac{1}{2}

Step-by-step explanation:

Given equation:

p^3=\dfrac{1}{8}

Cube root both sides:

\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}

\implies p= \sqrt[3]{\dfrac{1}{8}}

\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:

\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}

\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:

\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad 1^a=1:

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Rewrite 8 as 2³:

\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}

Simplify:

\implies p= \dfrac{1}{2^{\frac{3}{3}}}

\implies p= \dfrac{1}{2^{1}}

\implies p= \dfrac{1}{2}

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Answer:

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Step-by-step explanation:

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