1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2
Sector area = 0.5 * r^2 * angle (in radians)
so:
16 pi = 0.5 * r^2 * (90*pi/180)
r^2 = 16 pi /(0.25 pi) = 64
r = 8
therefore it would be d
Answer:
10000
Step-by-step explanation:
I hope this helps you
31 and x looking same arc
x=31
Answer:
23346.5
Step-by-step explanation:
add them together and divide by 2
