Question

(3.6x10^-5)÷(1.8x10^2) in standard form

Answer 1

Answer:

The standard form of $\left(3.6 \times 10^{-5}\right) \div\left(1.8 \times 10^{2}\right)$ is 0.0000002

Solution:

We have to write $\left(3.6 \times 10^{-5}\right) \div\left(1.8 \times 10^{2}\right)$ in standard form.

$=\frac{3.6 \times 10^{-5}}{1.8 \times 10^{2}}$

By dividing 3.6 by 1.8 we get “2”, hence the above equation becomes,

$=\frac{2 \times 10^{-5}}{10^{2}}$

We know that $\frac{x^{a}}{x^{b}}=x^{a-b}$

Therefore the above equation becomes,

$=2 \times 10^{-5-2}$

$=2 \times 10^{-7}$

since the exponent of 10 is negative , therefore 7 zeros are written on the left hand side of the number.

= 0.0000002

Hence the standard form of $\left(3.6 \times 10^{-5}\right) \div\left(1.8 \times 10^{2}\right)$ is 0.0000002

Answer 2

= (3.6/1.8) ( 10^-5/10^2)
= (2)(10^-7)
= 2/10^7
= 2/10000000
= .0000002