(2-7m^6)^2 i dont know this

1 answer:

7 0

(2-7m^6)^2

rearrange terms:

(2-7m^6)^2

(-7m^6 + 2)^2

expand the squares:

(-7m^6 + 2)^2

(-7m^6 + 2)(-7m^6 + 2)

distribute:

(-7m^6 + 2)(-7m^6 + 2)

-7(-7m^6 + 2) • m^6 + 2(-7m^6 + 2)

distribute:

-7(-7m^6 + 2) • m^6 + 2(-7m^6 + 2)

49m^12 - 14m^6 + 2(-7^6 +2)

distribute:

49m^12 - 14m^6 + 2(-7^6 +2)

49m^12 - 14m^6 - 14m^6 + 4

combine like terms:

49m^12 - 14m^6 - 14m^6 + 4

49m^12 - 28m^6 + 4

solution:

49m^12 - 28m^6 + 4

please mark brainliest and hope it helps

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Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
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-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$