32. 2x=64 can be written as x = 64/2. 64/2=32
3x+1+2x-2 = 64 3x+2x = 64-1+2 5x = 65 x = 13
Do you mean '16x^(4) -1 ; or ' (16 X 4 ) - 1' . Reason , the answers will be different. 16x^(4) - 1 = (2x)^(4) - 1^(4) = ((2x^(2) - 1^(2))((2x)^(2) + 1^(2)) = (2x - 1)(2x +1)((2x)^(2) + 1^(2)) = (2x - 1)(2x + 1)((2x)^(2) + 1) 2 & ( 16 X 4 ) - 1) = 64 - 1 = 63
(x - 2)(x + 2)(x2 - 2x + 4)(x2 + 2x + 4)
x2 - 2x + 65 = 0 Subtract 64 from each side: x2 - 2x + 1 = -64 (x - 1)2 = (i*8)2 where i is the imaginary square root of -1 x - 1 = +/- 8i x = 1 +/- 8i
michael jackson
find x when, 2x=128 2x=128 /2 /2 x=128 /2 x=64 then check your answer by doing the following if x=64, then what is 2x? 2(64)=? or 2 * 64=? or 2 x 64=? 2 times 64 equals 128!
32. 2x=64 can be written as x = 64/2. 64/2=32
-62
3x+1+2x-2 = 64 3x+2x = 64-1+2 5x = 65 x = 13
The answer is 32.
2X + 8^2 = 2X+64, for whatever value X. If the question is (2x+8)², the answer is 4x²+32x+64 The formula is (a+b)² = a² + 2ab + b². a=2x, b=8, a²=4x², b²=64
Do you mean '16x^(4) -1 ; or ' (16 X 4 ) - 1' . Reason , the answers will be different. 16x^(4) - 1 = (2x)^(4) - 1^(4) = ((2x^(2) - 1^(2))((2x)^(2) + 1^(2)) = (2x - 1)(2x +1)((2x)^(2) + 1^(2)) = (2x - 1)(2x + 1)((2x)^(2) + 1) 2 & ( 16 X 4 ) - 1) = 64 - 1 = 63
I think it's x^(3/2) x*x=x^2 sq.rt. x * sq. rt. x= x so x* sq. rt. x should be x^(1.5)...i think cuz each sq. rt. is worth half an exponent of 1. x^(1/2) = sq. rt. x hope this helps....
Two to the sixth power. 8 with an exponent of 2 equals 64 and 4 with an exponent of 3 equals 64
(x - 2)(x + 2)(x2 - 2x + 4)(x2 + 2x + 4)
x2 - 2x + 65 = 0 Subtract 64 from each side: x2 - 2x + 1 = -64 (x - 1)2 = (i*8)2 where i is the imaginary square root of -1 x - 1 = +/- 8i x = 1 +/- 8i