2007-01-03 05:26:01 · 14 answers · asked by electra 1 in Science & Mathematics ➔ Mathematics
5 2/3 = (5*3 + 2)/3 = 17/3
4 * 17/3 = 68/3 = 22 2/3
An alternate way that's easier to do in your head:
4 * 5 = 20
4 * 2/3 = 8/3 = 2 2/3
So 4 * 5 2/3 = 20 + 2 2/3 = 22 2/3
2007-01-03 05:28:31 · answer #1 · answered by Jim Burnell 6 · 2⤊ 0⤋
Math question 4 * 5 2/3
= 4(5 +2/3)
= 20 + 8/3
= 20 + 2 2/3
=22 2/3
2007-01-03 05:34:19 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
4 x 5 2/3 =
4 x 17/3 =
68/3 =
22 2/3
2007-01-03 10:44:28 · answer #3 · answered by David C 2 · 0⤊ 0⤋
First convert the mixed number 5 2/3 to a improper fraction
Multiply 3 times 5 and add 2
3 x 5 = 15 + 2 = 17/3
17 /3 is the improper fraction
Multiply 4 times 17/3 and divide by 3
4 x 17/3 =
68/3 =
22 2/3
22.66666666
- - - - - - -s-
2007-01-03 07:43:03 · answer #4 · answered by SAMUEL D 7 · 0⤊ 0⤋
4 * 5 2/3 = 4 * 17/3 = 68/3 = 22 2/3
QED
2007-01-03 07:01:54 · answer #5 · answered by CSUFGrad2006 5 · 0⤊ 0⤋
22 2/3
2007-01-03 05:29:14 · answer #6 · answered by strtat2 5 · 0⤊ 0⤋
22 2/3
2007-01-03 05:29:08 · answer #7 · answered by Tiff 5 · 0⤊ 0⤋
5 2/3=17/3
4*17/3=68/3=22 2/3
2007-01-03 06:29:53 · answer #8 · answered by yupchagee 7 · 0⤊ 0⤋
4 = 4/1
5 2/3 = 17/3
4/1 * 17/3 = 68/3
Ans: 22 2/3
2007-01-03 05:29:12 · answer #9 · answered by Tom :: Athier than Thou 6 · 0⤊ 0⤋
it somewhat is an algebra so it is so commonly used and significant part of Maths so the way of having an answer is on decrease than; 4+5(x+2) = 3(8x-3) 4 + 5x + 10 = 24x - 9 before everything, you should eliminate the bracket (multiply each and each numbers interior the bracket through exterior variety in turn) Then flow the organic numbers(ones with out x) to a minimum of one edge and the Xs to a distinctive. upload as much as a minimum of one variety the two edge 4 + 10 + 9 = 24x - 5x 23 = 19x 23/19 = x Then hit upon a x 23/19 = x (remember once you progression an variety to distinctive, exchange it into opposite e.g. + to -, - to +)
2016-11-26 00:50:40 · answer #10 · answered by Anonymous · 0⤊ 0⤋