Step-by-step explanation:
Your problem → 5y+5/2 / 25y-20/40y^2-32y
5y+5/2÷25y-20/40y^2-32y
=5⋅y+5/2÷25×y-20/40×y^2-32×y
=5y+5/2÷25×y-20/40×y^2-32y
=5y+5/2×1/25×y-20/40×y^2-32y
=5y+y/10-20/40×y^2-32y
=5y+y/10-y^2/2-32y
=5y×10+y-y^2×5-32y×10 ÷ 10
=50y+y-5y2-320y ÷ 10
= -269y-5y^2 / 10
Answer:
# = -2
Step-by-step explanation:
3 x # = -6
3 / -6 = -2
# = -2
Answer:
D) 53 Degrees.
Step-by-step explanation:
Things we need to establish beforehand: We know that Lines OZ and OX are equal because they are both radii of the circle. We can make an Iscoceles traingle by drawing a line between ZX. We know angle YZO and angle YXO is a right angle because YZ and XY are tangent to the circle. The Arc angle is the same angle as angle ZOX.
1) Find angles OZX and OXZ. these will be 26.5, because 180-127 is 53, which is the sum of the two angles. the two angles are the same, so divide 53 by 2.
2) Find Angles XZY and ZXY. We know that YZO is a right angle, and both XZY and OZX make up this right angle so XZY + OZX = 90. OZX is 26.5, so 90-26.5=XZY. XZY = ZXY, so both angles equal 63.5.
3) Now that we have two angles of triangle XYZ, we can find angle XYZ. 180-(XZY+ZXY)=XYZ, so (180-(63.5+63.5)=53. Angle XYZ=53.
1.04 ^ 1/12 = 0.086666666
or
1.04 ^ 1/12 = 13/150
Answer B. since it has 5x^4
