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7nadin3 [17]
4 years ago
14

Please help will give brainliest

Mathematics
2 answers:
elena55 [62]4 years ago
4 0

Answer:

I think is the last one

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Yanka [14]4 years ago
4 0

Step-by-step explanation:

they are exterior angles

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3 years ago
Find x. Round to the nearest tenth if needed.
alexdok [17]

Answer:

x = 16

Step-by-step explanation:

Use that the addition of all internal angles of a triangle must add up to 180, and the fact that the two given triangles are similar:

51 + 65 + 4 x = 180

combine and solve for "x"

116 + 4 x = 180

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3 years ago
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3 years ago
Identify the solution and graph of the given inequality. 34 &gt; 3(2 − x)
zloy xaker [14]

Answer:

x > 28/3

Step-by-step explanation:

34 > 3(2 − x)

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Sorry I don't know how to graph here, but I can graph it by writing the way it can be graphed.

It is in slope intercept form; y = mx + b

where mx is the slope and b is the y-intercept;

x>28/3; where 28/3 is the slope which means; at the y-intercept, go up 28 times and to the right 3 times

And b is 0; which means at 0 where the slope will be applied, go up 28 times and to the right 3 times.

Hope this helps!

Good luck.

3 0
4 years ago
Find the exact value of sin(cos^-1(4/5))
boyakko [2]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2762144

_______________


Let  \mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}


\mathsf{0\le \theta\le\pi,}  because that is the range of the inverse cosine funcition.


Also,

\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\&#10;\mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}


Square both sides and apply the fundamental trigonometric identity:

\mathsf{(5\,cos\,\theta)^2=4^2}\\\\&#10;\mathsf{5^2\,cos^2\,\theta=4^2}\\\\&#10;\mathsf{25\,cos^2\,\theta=16\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\&#10;\mathsf{25\cdot (1-sin^2\,\theta)=16}

\mathsf{25-25\,sin^2\,\theta=16}\\\\&#10;\mathsf{25-16=25\,sin^2\,\theta}\\\\&#10;\mathsf{9=25\,sin^2\,\theta}\\\\&#10;\mathsf{sin^2\,\theta=\dfrac{9}{25}}&#10;

\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}


But \mathsf{0\le \theta\le\pi,} which means \theta lies either in the 1st or the 2nd quadrant. So \mathsf{sin\,\theta} is a positive number:

\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\&#10;\therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>

3 0
3 years ago
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