Just need 37-45 answered

Problem 37)

Secant angle = [(big arc) - (small arc)]/2
30 = (x - 70)/2
2*30 = 2*(x - 70)/2
60 = x-70
60+70 = x-70+70
130 = x
x = 130

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Problem 38)

I'm assuming that is an 8 next to the y? Let me know if that assumption is correct or not. Thanks.

10^2 = (y+8)*8
100 = 8y+64
100-64 = 8y+64-64
36 = 8y
8y = 36
8y/8 = 36/8
y = 4.5

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Problem 39)

SA = 2*pi*r^2+2*pi*r*h
SA = 2*pi*8^2+2*pi*8*25
SA = 128pi+400pi
SA = 528pi
SA = 1658.76092109541
SA = 1658.76

The surface area is approximately 1658.76 square inches

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Problem 40)

I'm not exactly sure with this one. I don't see any connection forming so I don't see how you can use the law of detachment or the law of syllogism at all. Therefore I'm thinking that no valid conclusion can be drawn.

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Problem 41)

This is using the substitution property. AC is being substituted, or replaced, with BD

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Problem 42)

Arc EF = arc BD, so chord BD is congruent to chord EF.

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Problem 43)

AC is a perpendicular bisector to BD. So BD is cut exactly in half where BG = GD

GD = (1/2)*BD
GD = (1/2)*26
GD = 13 units

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Problem 44)

Let x be the unknown angle

"its measure is 4 times the measure of its complement" translates to this equation

x = 4*(90 - x)

Solve for x
x = 4*(90 - x)
x = 360 - 4x
5x = 360
x = 360/5
x = 72

So the original angle is 72 degrees

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Problem 45)

Reflecting over the x axis has you change any positive y coordinate to a negative y coordinate (or vice versa). So we have

A(0,4) ----> A'(0,-4)
B(5,0) ----> B'(5,0)
C(7,5) ----> C'(7,-5)

The two triangle has vertices at A'(0,-4), B'(5,0) and C'(7,-5)