well tonday was my first dady of school, and we already got a TON of homework in math. they were mostly review. but there was this 1... well 2 questions that totally stumped me.. so if anyone can help me that would b great. problem:
Insert grouping symbols in the expression 2*3^3+4 to produce the indicated values.
a. 62 b. 220 c. 4374 d. 279936
i already got the first two, so if anyone can help me on c and d that would b awesome. thanks!
2006-08-21 15:30:32 · 7 answers · asked by mmkay 2 in Science & Mathematics ➔ Mathematics
oh and if you dont get like what the thing is about, heres what i got for a:
2(3^3+4)
2(27+4)
2(31)
62.
so yeah..
2006-08-21 15:32:21 · update #1
c. 2*3^(3+4)
2(3^7)
4374
D. (2*3)^(3+4)
6^7
279936
2006-08-21 15:44:21 · answer #1 · answered by Anonymous · 0⤊ 0⤋
2*(3^(3+4)) = 2 * 3^7 = 2 * 2187 = 4374
(2*3)^(3+4) = 6^7 = 279936
2006-08-21 15:35:49 · answer #2 · answered by Anonymous · 0⤊ 0⤋
OK
c is 2*(3^(3+4)) = 2*3^7 = 2*2187 = 4,374
d is (2*3)^(3+4) = 6^7 = 279,936
Doug
2006-08-21 15:39:46 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
for c is 2*3^(3+4)
and d is (2*3)^(3+4)
Thats all..
2006-08-21 15:40:25 · answer #4 · answered by ArcherOmega 4 · 0⤊ 0⤋
c: 2 * (3^(3 + 4)) = 4 374
d: (2 * 3)^(3 + 4) = 279 936
2006-08-21 15:42:44 · answer #5 · answered by crissyll22 4 · 0⤊ 0⤋
2*3^(3+4)
2*3^7
I dont know what that is though
2006-08-21 15:39:31 · answer #6 · answered by Spelunking Spork 4 · 0⤊ 0⤋
I'm also eager to know that
2006-08-21 17:20:48 · answer #7 · answered by nima_iran_1985 3 · 0⤊ 0⤋