mk i have this problem
6y^2(2y^0)^2
i know it has to equal 24y^2
but how explain all the way please!
Thank you!
2007-09-07 16:55:31 · 7 answers · asked by Mae-Day 3 in Science & Mathematics ➔ Mathematics
6y^2(2y^0)^2
Note y^0 = 1
so
6y^2(2y^0)^2
= 6 y^2 (2^2)
= 6y^2 (4)
= 24y^2
2007-09-07 17:17:20 · answer #1 · answered by vlee1225 6 · 0⤊ 0⤋
First, any number to the zero power has a value of 1, so if you work on the parenthesis first, 2 y^0 is the same as 2 * 1 which is 2.
so the number in parenthesis is 2, then you square that to get 4,
then you multiply that 4 times 6y^2 which makes 24y^2.
2007-09-08 00:06:06 · answer #2 · answered by halrosser 5 · 0⤊ 0⤋
Remember that anything raised to the power of 0 (except for zero itself) is 1. So this is
6 y^2 (2*1)^2
This is 6 y^2 * 4 = (6*4)y^2 = 24 y^2
2007-09-07 23:59:44 · answer #3 · answered by Anonymous · 0⤊ 0⤋
any variable or constant raised to the power of zero(^0) is equal to 1
Thus,
6y^2(2y^0)^2
6y^2(4y^0)
6y^2[4(1)]
6y^2(4)
24y^2
Understand? Good luck!
2007-09-08 00:19:15 · answer #4 · answered by Anonymous · 0⤊ 0⤋
y^0 is equal to 1. So when you square the stuff in the ( ), all you are squaring is the 2.
2007-09-08 00:09:45 · answer #5 · answered by cattbarf 7 · 0⤊ 0⤋
i think you are cheating
but anyway it equals 24y^2
first term is 6y^2 second simplifies to 4
2007-09-08 00:03:44 · answer #6 · answered by Anonymous · 0⤊ 0⤋
its 7....
2007-09-07 23:59:26 · answer #7 · answered by leroy jenkins 2 · 0⤊ 1⤋