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Question-61537




Question Number 61537 by aanur last updated on 04/Jun/19
Commented by aanur last updated on 04/Jun/19
sir could you help me
sircouldyouhelpme
Commented by math1967 last updated on 04/Jun/19
xyz×(√(x^2 ×y^2 ×z^2 )) =4×9×16  x^2 ×y^2 ×z^2 =4×9×16⇒xyz=2×3×4
xyz×x2×y2×z2=4×9×16x2×y2×z2=4×9×16xyz=2×3×4
Answered by MJS last updated on 04/Jun/19
(1) ⇒ z=((16)/(x^2 y))  (2) ((4(√y))/( (√x)))=9  (3) ((16)/( (√(x^3 y))))=16    (2) ⇒ y=((81x)/(16)) ⇒ z=((256)/(81x^3 ))  (3) ((64)/(9x^2 ))=16 ⇒ x=(2/3) ⇒ y=((27)/8) ∧ z=((32)/3)
(1)z=16x2y(2)4yx=9(3)16x3y=16(2)y=81x16z=25681x3(3)649x2=16x=23y=278z=323
Answered by Kunal12588 last updated on 04/Jun/19
i) x(√(yz))=4  ii) y(√(xz))=9  iii) z(√(xy))=16  i×ii×iii  ⇒xyz(√(x^2 y^2 z^2 ))=2^6 ×3^2   ⇒(xyz)^2 =2^6 ×3^2   ⇒xyz=2^3 ×3=24  (i)^2   ⇒x^2 yz=4^2 =16  ⇒((x^2 yz)/(xyz))=((16)/(24))  ⇒x=(2/3)  (ii)^2   ⇒xy^2 z=81  ⇒y=((81)/(24))=((27)/8)  (iii)^2   ⇒xyz^2 =256  ⇒z=((256)/(24))=((32)/3)
i)xyz=4ii)yxz=9iii)zxy=16i×ii×iiixyzx2y2z2=26×32(xyz)2=26×32xyz=23×3=24(i)2x2yz=42=16x2yzxyz=1624x=23(ii)2xy2z=81y=8124=278(iii)2xyz2=256z=25624=323

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