Prove that 16^4–2^13–4^5 is divisible by 11

(16^4 -2^13 -4^5) mod 11 = ((2^4)^4 -2^13 -(2^2)^5) mod 11

= (2^16 -2^13 -2^10) mod 11

= (2^10 mod 11)·((2^6 -2^3 -1) mod 11)

= (1024 mod 11)·(55 mod 11)

= 1 · 0 = 0

When the sum shown above is divided by 11, its remainder is zero. Hence the sum is divisible by 11.

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25