3 years ago

Pls help 72 points given!

Mathematics

1 answer:

Inessa05 [86]3 years ago

5 0

Step-by-step explanation:

<h3>x = 1cm. ---- ( tangent property )</h3><h3>AB = 2x</h3><h3>AB = 2 × 1cm</h3><h3>AB = 2cm</h3><h3>___________________</h3><h2>FOLLOW ME</h2>

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2 years ago

Can you please help me find the missing segment to the triangle in the attached image? Thanks.

Keith_Richards [23]

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3 years ago

✓<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B19%20%2B%20%20%5Csqrt%7B30%20%2B%20%20%5Csqrt%7B32%20%2B%20x%20%20%7D%20%7D%

Setler79 [48]

We'll have to repeatedly square both sides of the equation, in order to get rid of the square roots. Squaring a first time yields

$19+\sqrt{30+\sqrt{32+x}}=25$

Move the 19 to the right hand side:

$\sqrt{30+\sqrt{32+x}}=6$

And square again:

$30+\sqrt{32+x}=36 \iff \sqrt{32+x}=6$

Square one last time:

$32+x=36 \iff x=36-32=4$

Let's check the solutions: all these squaring might have created external solutions:

$\sqrt{19+\sqrt{30+\sqrt{32+4}}}=\sqrt{19+\sqrt{30+6}}=\sqrt{19+6}=\sqrt{25}=5$

So, $x=4$ is a feasible solution.

8 0

3 years ago