Question 128

If $$\frac{a}{b}=\frac{4}{5}$$ and $$\frac{b}{c}=\frac{15}{16}$$, then $$\frac{18^{c^{2}}-7a^{2}}{45c^{2}+20a^{2}}$$ is equal to

Solution

Given that $$\frac{a}{b}=\frac{4}{5}$$ and $$\frac{b}{c}=\frac{15}{16}$$

we need to find the value of = $$\frac{18^{c^{2}}-7a^{2}}{45c^{2}+20a^{2}}$$

divide whole equation $$b^2$$

We will get ,

$$\frac{18 \frac{c}{b}^2 - 7 \frac{a}{b}^2 }{45 \frac{c}{b}^2 + 20 \frac{a}{b}^2}$$

$$\frac{18 \frac{16}{15}^2 - 7 \frac{4}{5}^2 }{45 \frac{16}{15}^2 + 20 \frac{4}{5}^2}$$

=$$\frac{1}{4}$$

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