If I simplify 4^(h-k) / 4^(h+k), is 4^(-2k) or 1^(-2k) the solution?
2007-07-20 08:20:47 · 5 answers · asked by jumba 1 in Science & Mathematics ➔ Mathematics
4^(h-k) / 4^(h+k),
=4^[(h-k) -(h+k)]
=4^(-2k)
2007-07-20 08:27:19 · answer #1 · answered by harry m 6 · 1⤊ 0⤋
you're able to try this a pair techniques. a million) y - 4 = a million - x ability y = 5 - x Substiute that into the 1st equation x + y = 4: x + (5 - x) = 4 5 = 4 No answer. 2) Rewrite y - 4 = a million- x to x + y = a million + 4 = 5 x + y won't be able to equivalent the two 4 and 5. No answer.
2016-10-22 04:29:50 · answer #2 · answered by serravalli 4 · 0⤊ 0⤋
4^(-2k) is true.
4^(h-k) / 4^(h+k)
4^h *4^-k * 4^-h * 4^-k
=(4^0 )* (4^-2k)
=4^(-2k)
=1/(2^4k)
2007-07-20 08:30:45 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
4^(-2k)
2007-07-20 08:41:44 · answer #4 · answered by punkass 2 · 0⤊ 0⤋
that is right answer!
2007-07-20 08:29:54 · answer #5 · answered by emma 1 · 0⤊ 0⤋