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Question 39In...

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Question 39

In case of inverse proportion, $\frac{a_{2}}{-} = \frac{b_{2}}{-}$

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Solution

In inverse proportion,
$\frac{a_{2}}{a_{1}} = \frac{b_{2}}{b_{1}}$
$\Rightarrow a_{2} b_{1} = a_{1} b_{2}$

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Q. Question 39

In the question given below, fill in the blank to make the statement true:

(a - b)^{2} +

= a^{2} - b^{2}

Q. Question 39

In the question given below, fill in the blank to make the statement true:

(a - b)^{2} +

= a^{2} - b^{2}

\begin{matrix} A & B A_{1} & B_{1} A_{2} & B_{2} \end{matrix}

If A and B are in inverse proportion, then which of the following is true?

Question 38

In direct proportion, $\frac{a_{1}}{b_{1}}$ ___ $\frac{a_{2}}{b_{2}}$ .

state True or False

The fourth proportion of

a^{2}, a b, b^{2}

\frac{b^{3}}{a}

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