Answer:
![x_{1} =-14\\\\x_{2} =6](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D-14%5C%5C%5C%5Cx_%7B2%7D%20%3D6)
Step-by-step explanation:
![x^{2} +8x-84=0\\\\x=\dfrac{-b \pm \sqrt{b^{2}-4ac } }{2a} \\\\a=1; \: \: b=8; \: \: c=-84\\\\x=\dfrac{-8 \pm \sqrt{8^{2}-4 \cdot 1 \cdot (-84) } }{2 \cdot 1}=\dfrac{-8 \pm \sqrt{400 } }{2}=\dfrac{-8 \pm20}{2} =-4 \pm 10\\\\x_{1} =-14\\\\x_{2} =6](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B8x-84%3D0%5C%5C%5C%5Cx%3D%5Cdfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D%20%5C%5C%5C%5Ca%3D1%3B%20%5C%3A%20%5C%3A%20b%3D8%3B%20%5C%3A%20%5C%3A%20c%3D-84%5C%5C%5C%5Cx%3D%5Cdfrac%7B-8%20%5Cpm%20%5Csqrt%7B8%5E%7B2%7D-4%20%5Ccdot%201%20%5Ccdot%20%28-84%29%20%7D%20%7D%7B2%20%5Ccdot%201%7D%3D%5Cdfrac%7B-8%20%5Cpm%20%5Csqrt%7B400%20%7D%20%7D%7B2%7D%3D%5Cdfrac%7B-8%20%5Cpm20%7D%7B2%7D%20%3D-4%20%5Cpm%2010%5C%5C%5C%5Cx_%7B1%7D%20%3D-14%5C%5C%5C%5Cx_%7B2%7D%20%3D6)
Answer:
![\begin{gathered} y=\frac{9}{6}x-9 \\ l=\frac{P}{2}-w \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D%5Cfrac%7B9%7D%7B6%7Dx-9%20%5C%5C%20l%3D%5Cfrac%7BP%7D%7B2%7D-w%20%5Cend%7Bgathered%7D)
Step-by-step explanation:
To solve the following equations, we need to isolate the variable asked using inverse operations to solve equations.
Remember that addition and subtraction are inverse operations, as multiplication and division.
Therefore,
![\begin{gathered} a\text{. Solve for y} \\ 6y=9x-54 \\ y=\frac{9}{6}x-\frac{54}{6} \\ y=\frac{9}{6}x-9 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%5Ctext%7B.%20Solve%20for%20y%7D%20%5C%5C%206y%3D9x-54%20%5C%5C%20y%3D%5Cfrac%7B9%7D%7B6%7Dx-%5Cfrac%7B54%7D%7B6%7D%20%5C%5C%20y%3D%5Cfrac%7B9%7D%7B6%7Dx-9%20%5Cend%7Bgathered%7D)
<h3>
Answer: B) 49.22</h3>
========================================================
Work Shown:
cos(angle) = adjacent/hypotenuse
cos(43) = 36/x
x*cos(43) = 36
x = 36/cos(43)
x = 49.2237885995494
x = 49.22
Your calculator needs to be in degree mode.
What are the answer of your list ?
Answer:
Step-by-step explanation:
<h2>can somebody help please<em>∈∈</em></h2>