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Answers:</h3>
- Problem 5) x = 4
- Problem 6) y = 8
- Problem 7) z = 4
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Explanations:
Problem 5)
Assuming HEFG is a parallelogram, this means the opposite sides are the same length.
EH = GF
x-3 = 4x-15
x-4x = -15+3
-3x = -12
x = -12/(-3)
x = 4
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Problem 6)
We'll use the same idea from problem 5. The opposite sides EF and HG are congruent
EF = HG
3y = 3x+12
3y = 3*4+12
3y = 24
y = 24/3
y = 8
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Problem 7)
For any parallelogram, the diagonals always cut each other in half.
Therefore HK = 2z+3 is exactly half that of segment HF = 22
Put another way, HF is twice as long as HK
2*(HK) = HF
2*(2z+3) = 22
4z+6 = 22
4z = 22-6
4z = 16
z = 16/4
z = 4
Answer:
Hope It Help
Step-by-step explanation:
Brainliest please
<span>The domain of the function can be obtained by pluging the values of the range into f(x), i.e domain = {k^2 + 2k + 1 = 25, k^2 + 2k + 1 = 64} = {k^2 + 2k - 24 = 0, k^2 + 2k - 63 = 0}. Solving the two quadratic equations, we have that the range is {-9, -6, 4, 7}.I hope that my answer is helpful! Let me know if you need something more :)</span>
C is the hypotenuse, so the pythagorean theorem used would be: 15²=13²+6². 225=169+36, but 169+36 does not equal to 225, therefore the triangle is not a right triangle
