Prove that 5^31–5^29 is divisible by 100.

5^31-5^29

=5^29(5^2-1)

=5^29(25-1)

=5^29(24) ...a

100=5^2x4 ...b

Divide a by b

5^29(24) / 5^2 x 4

=5^27 x 6

This, 5^31-5^29 is divisible by 100.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

the formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. if a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.