2007-08-01 20:15:24 · 7 answers · asked by paavni 2 in Science & Mathematics ➔ Mathematics
= 5 [ 2 + 3 ] ³ ]^(1/4)
= (5) 5^(¾)
= 5^(7/4)
2007-08-01 21:18:25 · answer #1 · answered by Como 7 · 1⤊ 0⤋
5[[8 to the power 1/3 + 27 to the power 1/3] to the power 3]to the power 1/4 = 5[[[2to the power 3]to the power 1/3 + [3to the power 3]to the power 1/3]to the power 3]to the power 1/4 = 5[[2+3]to the power 3]to the power 1/4 =5[[5]to the power 3]to the power 1/4 = 5[125]to the power 1/4 =5[3.343701525=16.71850762
i guess the question should be : [5[[8 to the power 1/3 + 27 to the power 1/3]to the power 3]]to the power 1/4 , ie . less of one bracket [ ]. the solution for [5[[8 to the power 1/3 + 27 to the power 1/3]to the power 3]]to the power 1/4 is :
[5[[8 to the power 1/3 + 27 to the power 1/3]to the power 3]]to the power 1/4 = [5[[2to the power 3]to the power 1/3 + [3to the power 3]to the power 1/3]to the power 3]]to the power 1/4 = [5[[2+3]to the power 3]]to the power 1/4 =[5[[5]to the power 3]]to the power 1/4 = [5 to the power 4]to the power 1/4 = 5
2007-08-02 20:39:13 · answer #2 · answered by follower 1 · 0⤊ 0⤋
5 [ (8^1/3 +27^1/3 ) ^3 ] ^1/4
5[ (2+3) ^3 ] ^1/4
5 (5^3)^1/4
5 (5^3/4)
As the base is same ie 5 so we can add the powers
5^ (1+3/4)
5^(7/4)
2007-08-01 20:19:05 · answer #3 · answered by sweet n simple 5 · 1⤊ 0⤋
5[(8^(1/3)+27^(1/3))^(3)]^(1/4...
=5[(2+3)^3]^(1/4)
=5[5^(3/4)]
=5^(1+3/4)
=5^(7/4)
2007-08-01 20:22:15 · answer #4 · answered by Anonymous · 0⤊ 0⤋
5[(8^(1/3)+27^(1/3))^(3)]^(1/4...
=5[(2+3)^3]^(1/4)
=5[5^(3/4)]
=5^(1+3/4)
=5^(7/4)
2007-08-01 20:19:21 · answer #5 · answered by niki einstien 2 · 0⤊ 0⤋
5^(7/4)
2007-08-01 20:23:09 · answer #6 · answered by Anonymous · 0⤊ 1⤋
5[(8^⅓ + 27^⅓)^3]^¼
5[(2 + 3)^3]^¼
5[(5)^3]^¼
5(5)^¾
5(3∙343 701 525...)
16∙718 507 62...
2007-08-01 22:32:19 · answer #7 · answered by Sparks 6 · 0⤊ 1⤋