Answer:
what probortation?
Step-by-step explanation:
Answer:
A) x degrees + 65 degrees = 180 degrees
B) x degrees = z degrees
C) y degrees = 65 degrees
Step-by-step explanation:
Refer the attached figure
So,![\angle AOB = \angle COD (\text{Vertically opposite angles})](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%20%5Cangle%20COD%20%20%28%5Ctext%7BVertically%20opposite%20angles%7D%29)
![\angle AOB = 65^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%2065%5E%7B%5Ccirc%7D)
![\angle COD = y](https://tex.z-dn.net/?f=%5Cangle%20COD%20%3D%20%20y)
So, ![y=65^{\circ}](https://tex.z-dn.net/?f=y%3D65%5E%7B%5Ccirc%7D)
![\angle AOB + \angle BOC = 180^{\circ} (\text{Linear pair})](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%2B%20%5Cangle%20BOC%20%3D%20180%5E%7B%5Ccirc%7D%20%28%5Ctext%7BLinear%20pair%7D%29)
![\angle AOB = 65^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%2065%5E%7B%5Ccirc%7D)
![\angle BOC = x](https://tex.z-dn.net/?f=%5Cangle%20BOC%20%3D%20x)
So, ![x+65^{\circ}=180^{\circ}](https://tex.z-dn.net/?f=x%2B65%5E%7B%5Ccirc%7D%3D180%5E%7B%5Ccirc%7D)
![\angle BOC = \angle AOD (\text{Vertically opposite angles})](https://tex.z-dn.net/?f=%5Cangle%20BOC%20%3D%20%5Cangle%20AOD%20%28%5Ctext%7BVertically%20opposite%20angles%7D%29)
So, x = z
So, Option A ,B and C are true for the values of x, y, and z
Since the square root of
x=√x
and the cube root of
x=3√x
and their product
=√x × 3√x
= x1/2 × x1/3 [Since, n√a =a 1/n]
=x1/2+1/3 [Since, a^m ×a^n =a^m+n]
=x^3+2 /6
= x 5/6
Therefore, required product is
=√6x5