If 6 a3 b- 12 a2 b2- k and 48 a b3 are in proportion- then express k in terms of a and b-

It #160;is #160;given #160;that #160;6a3b,12a2b2, #160;k #160;and #160;48ab3 #160;are #160;in #160;proportion.Thus, #160;we #160;have:6a ...

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Standard IX

Mathematics

Direct Proportionality

If 6 a3 b, 12...

Question

If 6a³b, 12a²b², k and 48ab³ are in proportion, then express k in terms of a and b.

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Solution

$It is given that 6 a^{3} b, 12 a^{2} b^{2}, k and 48 a b^{3} are in proportion . Thus, we have : \frac{6 a^{3} b}{12 a^{2} b^{2}} = \frac{k}{48 a b^{3}}$ ^{$\Rightarrow k = \frac{6 a^{3} b \times 48 a b^{3}}{12 a^{2} b^{2}}$

$\Rightarrow k = 24 a^{2} b^{2}$}

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If 6a3b, 12a2b2, k and 48ab3 are in proportion, then express k in terms of a and b.

It is given that 6a3b,12a2b2, k and 48ab3 are in proportion.Thus, we have:6a3b12a2b2=k48ab3 ⇒k=6a3b×48ab312a2b2 ⇒k =24a2b2

If 6a³b, 12a²b², k and 48ab³ are in proportion, then express k in terms of a and b.

$It is given that 6 a^{3} b, 12 a^{2} b^{2}, k and 48 a b^{3} are in proportion . Thus, we have : \frac{6 a^{3} b}{12 a^{2} b^{2}} = \frac{k}{48 a b^{3}}$ ^{$\Rightarrow k = \frac{6 a^{3} b \times 48 a b^{3}}{12 a^{2} b^{2}}$

$\Rightarrow k = 24 a^{2} b^{2}$}