Question

If $\frac{\sqrt{954+\sqrt{\sqrt{484}+\sqrt{704+625}}}}{0.00155}=\frac{\sqrt{0.0004}}{{\left(0.009\right)}^{2}x}$,
then $x$ is:

• A. 123.456
• B. 0.01234
• C. 12.34
• D. 12345.67

Answer 1

The correct option is B 0.01234

$\frac{\sqrt{954+\sqrt{\sqrt{484}+\sqrt{704+25}}}}{0.00155}=\frac{\sqrt{0.0004}}{{\left(0.009\right)}^{2}x}$
= $\frac{\sqrt{954+\sqrt{\sqrt{484}+\sqrt{729}}}}{0.00155}=\frac{\sqrt{0.0004}}{{\left(0.009\right)}^{2}x}$
= $\frac{\sqrt{954+\sqrt{49}}}{0.00155}=\frac{\sqrt{0.0004}}{{\left(0.009\right)}^{2}x}$
= $\frac{\sqrt{954+7}}{0.00155}=\frac{\sqrt{0.0004}}{{\left(0.009\right)}^{2}x}$
= $\frac{31}{0.00155}=\frac{0.02}{{\left(0.009\right)}^{2}x}$
= $x=\frac{0.02\times 0.00155}{31\times (0.009{)}^2}$
= $x=0.012345$