There are two numbers whose $400^{\text{th}}$ powers are equal to $9^{1000}$. In other words, there are two numbers that can replace $\blacksquare$ in the equation $$\blacksquare^{400} = 9^{1000},$$making the equation true.

Find the smaller of the two numbers.

± 243

Work:

x^400 = 9^1000

so the GCF of 400, 1000  = 200

(x^2)^200 = (9^5)^200

(x^2) = 9^5      

(x^2) = (3^2)^2

x^2  = 3^10    

x = ± √[3^10]  =   ± [ 3] ^(10/2)  =  ± [3]^5  =  ± 243

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation:

answer: both chords and diameters have two endpoints on a circle. diameters must intersect the center of a circle.