2007-08-30 13:17:17 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
= 9 ² - 7 x 4
= 81 - 28
= 53
2007-09-03 11:27:16 · answer #1 · answered by Como 7 · 0⤊ 0⤋
(3x)^2 - 7y^2 when x=3 and y=2?
(3x)² - 7y²
Substitute the values of 'x' & 'y' in the above equation
{3*(3)}² - 7*(2)²
(9² - 7*4)
81 -28 = 53. . . . Ans
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2007-08-30 21:24:36 · answer #2 · answered by Joymash 6 · 0⤊ 0⤋
Substitute the values in for x and y so you end up with (3*3)^2-7*2^2. Then, remembering your order of operations, multiply inside the parenthesis giving you 9^2-7*2^2. Next come exponents giving you 81 [9*9 or 9^2] - 7*4 [2*2 or 2^2]. Now multiply giving you 81-28. Lastly subtract leaving you with 53.
2007-08-30 20:28:22 · answer #3 · answered by mikethetomato 1 · 0⤊ 0⤋
Plug in the numbers and simplify
(3*3)^2 - 7(2^2) =
9^2 - 7(4) =
81 - 28 = 53
2007-08-30 20:24:17 · answer #4 · answered by saberhilt 4 · 2⤊ 0⤋
(3x)^2 - 7y^2
(3(3))^2 - 7(2)^2
9^2 - 7(4)
81 - 28
= 53
2007-08-30 20:24:40 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(3(3))^2 - 7(2)^2
= 9^2 - 28
= 81 - 28
= 53
2007-08-30 20:24:55 · answer #6 · answered by frank 7 · 0⤊ 0⤋
Plug them in, plug them in. (3*3)^2=81
7*(2)^2= 28. So 81-28=53
2007-08-30 20:24:50 · answer #7 · answered by cattbarf 7 · 0⤊ 0⤋
(3x)^2-7y^2
(3x3)^2-7x2^2
9^2-7x4
81-28
53
yep sounds about right.
i checked my work.
go with this answer; i guarantee you wont regret it!
im 13 and im in math 1-4!!!
2007-08-30 20:28:57 · answer #8 · answered by Sky Chord 2 · 0⤊ 0⤋
9^2 - 7(4) = 0
81 - 28 = 0
53
2007-08-30 20:26:42 · answer #9 · answered by thehazmater 2 · 0⤊ 0⤋
[(3)(3)]^2-7(2)^2 =
6^2 -7(4)=
36-28=
8
i think. .
2007-08-30 20:22:45 · answer #10 · answered by Anonymous · 0⤊ 0⤋