If A = 6 and B = 5, what is the value of (A^2 - B ^2)^2?

D. 121

If A equals 6 and B equals 5...

(6^2 - 5^2)^2

((6 x 6) - (5 x 5))^2

(36 - 25)^2

11^2

11 x 11

121

Answer is D
(6^2 - 5^2)^2
= (36-25)^2
= 11^2
= 121

If A = 6 and B = 5, what is the value of
(A^2 - B ^2)^2
= 11^2 (if A = 6 and B = 5)
Answer choice:
D. 121

Answer is D because [A^2 - B^2]^2 = [6^2 - 5^2]^2 = [(6-5)(6+5)]^2 = 11^2 = 121.

D.121

Substitute a = 6 and b = 5 into the expression to get final answer of:
D. 121

very much D.

You have your formula, (A^2-B^2)^2
All you have to do is plug in your numbers.
(6^2-5^2)^2=
(36-25)^2=
11^2=
121
So the answer is D

You have your formula, (A^2-B^2)^2
All you have to do is plug in your numbers.
(6^2-5^2)^2=
(36-25)^2=
11^2=
121
So the answer is D.

D.121

(A^2 - B ^2)^2
= 11^2 (if A = 6 and B = 5)
Answer choice:
D. 121

121 is the answer

(6^2 - 5^2)^2
(36 - 25)^2
(11)^2
121

(A^2 - B^2)^2 = (6^2 - 5^2)^2

(A^2 - B^2)^2 = (36 - 25)^2

(A^2 - B^2)^2 = (11)^2

(A^2 - B^2)^2 = 11^2

(A^2 - B^2)^2 = 121

Therefore, the answer is D.

You have :
----------------

Let P = ( A^2 - B^2 )^2 with A = 6 and B = 5

P = ( 6^2 - 5^2 )^2

P = ( 36 - 25 )^2

P = ( 11 )^2 = 121 <---------------------

D. 121

f A equals 6 and B equals 5...
(6^2 - 5^2)^2
((6 x 6) - (5 x 5))^2
(36 - 25)^2
11^2
11 x 11
121

(((A^2)-(B^2))^2)
((6^2)-(5^2)) ^2
(36-25)^2
(11)^2
121

d. 121

D. 121

d

D.121

(36-25)^2=121

121. Use the order of operations. You'll do what's in the parenthesis first, starting with the squares and then subtraction. so 36 - 25 = 11, then square 11, so 11*11 = 121!

):/)7/)/)/6/6/

121

D. 121

D

You have :
--------------

P = f(A,B) = ( A^2 - B^2 )^1/2

P = [ 6^2 - 5^2 ]^2

P = [ 36 - 25 ]^2 = [ 11 ]^2 = 121 <----------------

OK, let's get tricky (and do it the complicated way)

a^2 - b^2 is a difference of squares
the factors are (a+b)(a-b)

therefore, we can rewrite as
[(a+b)(a-b)]^2
a=6 and b=5
gives us
a+b = 11
a-b = 1
their product (a+b)(a-b) = 11 * 1 = 11
this gets squared
11^2 = 11 * 11 = ...

You know that 11*11 is going to be bigger than 10*10,
and that is equal to 100.
Therefore, without doing the calculation, you can eliminate any choice that is less than 100.

[(6)^2-(5)^2]^2
[36-25]^2
11^2
121

Don't forget to solve what's in the parenthesis first.

(6^2-5^2)^2
((6x6)-(5x5))^2
(36-25)^2
36-25=11
(11)^2
11x11=121

D = 121

11

121

121

121

(A^2 - B ^2)^2, here if you substitute A and B by 6 and 5, you get (36-25)^2= 121 which is D in your option. However, for bigger values, you could write (A^2 - B ^2)^2= {(A+B)(A-B)}^2= (A+B)^2.(A-B)^2

its 121

(6^2 - 5^2)^2
= (36 - 25)^2
= 11^2
= 121

=(6^2 - 5^2)^2
=(36 - 25)^2
= 11^2
=121 answer//

Answer is D. 121..//

(A^2 - B ^2)^2
(36 - 25)^2
11^2 = 121

((6^2) - (5^2))^2
(36-25)^2
11^2
=121
so the answer will be d

(A^2- B^2)^2 = (( A +B) ( A-B) )^2 = (( 6+5) ( 6-5))^2 = (11*1)^2 = 11^2 = 121

ANSWER is D

D

D

It's 121, and I figured that out in my head.

Answer is D .

**A=6 and B=5**

(A^2-B^2)^2

= (6^2-5^2)^2

=(36-25)^2

=(11)^2

=D. 121

121

121

D. 121

D. 121

One Hundred Twenty One. :) Letter D.

Answer is D.121

36-25= 11
11 square= 121

(6^2 - 5^2)^2

(36 - 25)^2

(11)^2

D. 121

D

Choice D. 121 is correct.

121 just plug the numbers to the formula

121

(6^2 - 5^2)^2
(36-25)^2
(11)^2
121

D. 121