Answer:
PQR and JK
Step-by-step explanation:
ΔGHJ ≈ ΔPQR by the Transitive Property of Congruence. GH ≈ JK since corresponding parts of congruent figures are congruent.
Pls mark as brainliest. I would really appreciate it if you did.
Note: I'm afraid I can't put the work for the answers, for lack of time =/
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1. 0 or -3
2. 0 or 4
3. -2 or 3
4. 0 or 1/2
5. -2 or 3
6. 0 or 5
7. 0 or 7
8. 0 or -2
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9. Substitute 0 for h in "h = -16x^2 + 16x" then solve for x
0 = -16x^2 + 16x
<span>Subtract -16x^2+16x from both sides.
</span><span>0−<span>(<span><span>−<span>16x2</span></span>+16x</span>)</span></span>=<span><span><span>−<span>16<span>x2</span></span></span>+<span>16x</span></span>−<span>(<span><span>−<span>16x2</span></span>+16x</span>)
</span></span><span><span>16<span>x2</span></span>−<span>16x</span></span>=<span>0
</span>
Factor the left side of the equation
<span><span>16x</span><span>(<span>x−1</span>)</span></span>=<span>0
</span>
Set factors to equal 0
<span><span>16x</span>=<span><span><span>0<span> or </span></span>x</span>−1</span></span>=<span>0
</span>x = 0 or x = 1
Answer: 
Step-by-step explanation:
Properties of logarithm:
Consider,
![2\log x-\log y+2 \log z\\\\ =\log x^2-\log y+\log z^2\ \ \ \ \text{[By (i)]}\\\\= \log x^2+\log z^2-\log y\\\\=\log(x^2z^2)-\log y\ \ \ \ [\text{By } (ii) ]\\\\=\log(\dfrac{x^2z^2}{y}) \ \ \ \ [\text{By (iii)}]](https://tex.z-dn.net/?f=2%5Clog%20x-%5Clog%20y%2B2%20%5Clog%20z%5C%5C%5C%5C%20%3D%5Clog%20x%5E2-%5Clog%20y%2B%5Clog%20z%5E2%5C%20%5C%20%5C%20%5C%20%5Ctext%7B%5BBy%20%28i%29%5D%7D%5C%5C%5C%5C%3D%20%5Clog%20x%5E2%2B%5Clog%20z%5E2-%5Clog%20y%5C%5C%5C%5C%3D%5Clog%28x%5E2z%5E2%29-%5Clog%20y%5C%20%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20%7D%20%28ii%29%20%5D%5C%5C%5C%5C%3D%5Clog%28%5Cdfrac%7Bx%5E2z%5E2%7D%7By%7D%29%20%20%20%20%5C%20%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20%28iii%29%7D%5D)
Nine thousand, two hundred.
