y2= y squared
(2y)2= (2y) squared
please help me =(
2007-05-14 14:36:55 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
na
y^2 + 8^2
cos (2y)^2 is 4y^2
2007-05-14 14:43:35 · answer #1 · answered by vivien h 3 · 1⤊ 0⤋
So you want to solve for y?
(2y)2 = 2y*2y = 4y2
You have y2 + 64 = 4y2
You subtract y2 from both sides and get
64 = 3y2
You divide both sides by 3 and get
21.33 = y2
Take the square root of both sides and get
y= 4.62
2007-05-14 14:48:10 · answer #2 · answered by Angie 1 · 0⤊ 0⤋
y=8
y2 + 64 = (2y)2 subtract the y2 from the left
64=y2now square root the entire equation
y=8
2007-05-14 14:41:35 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Subtr y2
64 = 4y^2 - y^2
3y^2 = 64
y^2 = 64/3
y = + - sq rt (64/3)
y = + - 8 / sq rt 3 Rationalize
y = + - 8 sq rt 3 / 3
2007-05-14 14:40:19 · answer #4 · answered by richardwptljc 6 · 1⤊ 0⤋
y² + 64 = (2y)²
y² + 64 = 4y²
3y² = 64
y² = 64/3
y = 8/√3
or
Multiply thru by √3 / √3
y = (8√3)/3, y = -(8√3)/3
.
2007-05-14 14:45:33 · answer #5 · answered by Robert L 7 · 1⤊ 0⤋
y^2+64=(2y)^2
y^2+64=4y^2
64=4y^2-y^2
3y^2=64
y^2=64/3
y=root of 64/3
2007-05-14 15:14:04 · answer #6 · answered by sam 3 · 0⤊ 0⤋
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2016-12-29 04:45:27 · answer #7 · answered by Anonymous · 0⤊ 0⤋