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Norma-Jean [14]
3 years ago
5

Please help me !!

Mathematics
2 answers:
Brut [27]3 years ago
8 0
The answers to your questions are 1.D, and 2.C
irinina [24]3 years ago
7 0
1)~-8x-(8x)=-48\iff -16x=-48\iff 16x=48\iff\\\\\x=\dfrac{48}{16}\iff \boxed{x=3}

2)~y-y_p=m(x-x_p)\Longrightarrow\boxed{y-3=-2(x-2)}
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50 points. Please explain each step​
3241004551 [841]
<h3>Sin\theta_{1}=\frac{3\sqrt{21}}{17}</h3>

Solution given:

Cos\theta_{1}=\frac{10}{17}

\frac{adjacent}{hypotenuse}=\frac{10}{17}

equating corresponding value

we get

adjacent=10

hypotenuse=17

perpendicular=x

now

by using Pythagoras law

Hypotenuse ²=perpendicular²+adjacent ²

substituting value

17²=x²+10²

17²-10²=x²

x²=17²-10²

x²=189

doing square root

\sqrt{x²}=\sqrt{189}

x=3\sqrt{21}

now

In I Quadrant sin angle is positive

Sin\theta_{1}=\frac{perpendicular}{hypotenuse}

<h3>Sin\theta_{1}=\frac{3\sqrt{21}}{17}</h3>
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