Question

if the square root of 4096 is 64 then find the value of the sum of the square root of 4096 and 40.96

Answer 1

Given that the
$\sqrt{4096} = 64$

then

$\sqrt{40.96} = 6.4$

Then adding the square
root of 4096 & 40.96

= 64 + 6.4
= 70.4

${\mathfrak{hope\:it\:helps}}$

Answer 2

Answer:

The sum of the square root of 4096 and 40.96 is 70.4.

Step-by-step explanation:

Given, square root of 4096 is 64

So,

$\sqrt{4096} = 64$

Now we want to find the sum of the square root of 4096 and 40.96.

Therefore question is

$( \sqrt{4096} + \sqrt{40.96} ) =$?

Now we know,

$\sqrt{4096} = 64$

and

$\sqrt{40.96} \\ = \sqrt{ \frac{4096}{100} } \\ = \frac{ \sqrt{4096} }{ \sqrt{100} } \\ = \frac{64}{10} \\ = 6.4$

So,

$( \sqrt{4096} + \sqrt{40.96} ) = 64 + 6.4 = 70.4$

Therefore Required value is 70.4.

This is a problem of Power of indices.

Some important formulas of Power of

Some important formulas of Power ofindices,

${x}^{0} = 1 \\ {x}^{1} = x \\ {x}^{a} \times {x}^{b} = {x}^{a + b} \\ \frac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b} \\ {x}^{ {y}^{a} } = {x}^{ya} \\ {x}^{ - 1} = \frac{1}{x} \\ {x}^{a} \times {y}^{a} = {(xy)}^{a}$