If a=b²x , b=c²y , c=a²z, find the value of xyz
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Answered by
8
So xyz = 1/abc.
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Answered by
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Heyy here is U r Ans :=>>>
1. solution :=>
a = b²x
x = a/b², if b ≠ 0
b = c²y
y = b/c², if c ≠ 0
c = a²z
z = c/a², if a ≠ 0
For a ≠ 0, b ≠ 0, c ≠ 0,
xyz
= (a/b²)(b/c²)(c/a²)
= abc/(a²b²c²)
= 1/(abc)
If either a, b, or c is equal to zero, then all three are zero. In that case xyz may have any real value.
2. solution:=>
A = b^2 x
divide both sides by b^2
A / b^2 = x
b= c^2 y
divide both sides by c^2
b / c^2 = y
c = a^2 z
divide both sides by a^2
c / a^2 = z
x y z = ( A / b^2) ( b / c^2) ( c / a^2)
xyz = A b c / (a^2 b^2 c^2)
xyz = A / (a^2 b c)
Did you mean 'a' instead of 'A?' If so:
xyz = a b c / (a^2 b^2 c^2)
= 1 / (abc)
I hope this helps U :-)
1. solution :=>
a = b²x
x = a/b², if b ≠ 0
b = c²y
y = b/c², if c ≠ 0
c = a²z
z = c/a², if a ≠ 0
For a ≠ 0, b ≠ 0, c ≠ 0,
xyz
= (a/b²)(b/c²)(c/a²)
= abc/(a²b²c²)
= 1/(abc)
If either a, b, or c is equal to zero, then all three are zero. In that case xyz may have any real value.
2. solution:=>
A = b^2 x
divide both sides by b^2
A / b^2 = x
b= c^2 y
divide both sides by c^2
b / c^2 = y
c = a^2 z
divide both sides by a^2
c / a^2 = z
x y z = ( A / b^2) ( b / c^2) ( c / a^2)
xyz = A b c / (a^2 b^2 c^2)
xyz = A / (a^2 b c)
Did you mean 'a' instead of 'A?' If so:
xyz = a b c / (a^2 b^2 c^2)
= 1 / (abc)
I hope this helps U :-)
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