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2006-06-18 20:12:14 · 10 answers · asked by Diva 1 in Science & Mathematics Mathematics

10 answers

x√x=729
x³=729²=531,441
x=(531441)^(1/3)=81

2006-06-18 20:17:18 · answer #1 · answered by Pascal 7 · 1 0

X*x=243*3=729
sqrt=27

2006-06-18 20:20:09 · answer #2 · answered by Anonymous · 0 0

X=81

2006-06-18 20:30:54 · answer #3 · answered by Superstar 5 · 0 0

I got same answer as above (i.e. x = 81).

My working:

Problem: (Square root x)/3 = 243/x

Square (^2) both sides to get rid of the square root:
x/9 = 243^2/x^2
x/9 = 59049/x^2

Multiply both sides by 9 to get rid of fraction on LHS:
x = 59049/x^2 * 9
x = 531441/x^2

Multiply both sides by x^2 (x squared) to get x's on same side noting that x^2's on RHS are cancelled out:
x^3 = 531441

Finally, take the cube root of both sides to get the value of x:
Hence, x = 81.

2006-06-18 20:25:29 · answer #4 · answered by rouge7487 1 · 0 0

Is it [sq rt. (x)] / 3 Or sq. rt ( x / 3 ) ??

Taking it to be 1st,

x * sqrt. x = 243 * 3 = 81 * 9

Hence, x = 81 [ 9 is sq.rt of 81]



Taking it to be 2nd
x * sq.rt x = 243 * sqrt. 3
x^3 = 243 * 243 * 3 squaring
x = cube root (177147)

2006-06-18 20:45:31 · answer #5 · answered by nayanmange 4 · 0 0

sqrt(x)/3 = 243/x
cross multiply
xsqrt(x) = 729
sqrt(x^2 * x) = 729
sqrt(x^3) = 729
square both sides
x^3 = 531441
cbrt both sides
x = 81

ANS : 81

2006-06-19 05:42:31 · answer #6 · answered by Sherman81 6 · 0 0

81

2006-06-18 20:34:09 · answer #7 · answered by rajesh_583 1 · 0 0

sq. root 8 + squareroot 18 = X^2 2(2)^a million/2 + 3(2)^a million/2 = X^2 5(2)^a million/2 = X^2 X = (50)^a million/4 = 2.659 yet i think of you have asked the question incorrect. it rather is probably, sq. root of 18 + sq. root of 8 = sq. root of X oppsed to sq. of X. wherein case the respond would be 50

2016-12-13 17:12:34 · answer #8 · answered by sory 3 · 0 0

Squaring both side & then cross multiply.
U will get 27 answer

2006-06-19 00:15:44 · answer #9 · answered by Jatin 2 · 0 0

x/3=3^5/x
x^2=3^5*
x^2=3^6
x^2=729
x=27

2006-06-19 00:49:08 · answer #10 · answered by Anonymous · 0 0

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