Question

how to find the sqrt of 12.57

Answer 1

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Answer 2

without a calculator?

Answer 3

taylor series?

Answer 4

Assuming you mean with a calculator, just use the square root function:
\[\sqrt{12.57}\] = 3.54541958
= 3.55

Answer 5

wow unkle is typing alot

Answer 6

\[
\sqrt{12.57}\\
=(16-3.43)^{1/2}\\
=4\left(1-\frac{3.43}{16}\right)^{1/2}\\
=4\left[1+\tfrac12\Big(\frac{-3.43}{16}\Big)+\frac{\tfrac12(\tfrac12-1)}{2!}\Big(\frac{-3.43}{16}\Big)^2+\mathcal O\big((\tfrac{-3.43}{16})^3\big)\right]\\
=4\left[1+\Big(\frac{-3.43}{32}\Big)-\frac{\tfrac14}{2}\Big(\frac{-3.43}{16}\Big)^2+\mathcal O\big((\tfrac{-3.43}{16})^3\big)\right]
\\\approx4\times\left(1-\frac{3.43}{32}-\tfrac18\times\left(\frac{3.43}{16}\right)^2\right)
\\=3.548\]

Answer 7

holy mcdonald!!!

Answer 8

close enough

Answer 9

uncle from sqrt(12.57) where did you get (16 - 3.43)^1/2

Answer 10

well i know the square root of 16,

Answer 11

isnt it 4

Answer 12

16-3.43 = 12.57. He did it as he knew the sqrt of 16.

Answer 13

yes

Answer 14

ohhh

Answer 15

then factor out the four , and apply a binomial series

Answer 16

Alternate way:
Let \(\Large f(x) = \sqrt x\)
We need for x=12.57
As we now....
\[\Large f(x+ \Delta x) = f(x) + f' (x) [ \Delta x]\]
\[\Large f(16-3.43) = f(16) + \frac{1}{2 \sqrt 16 } \times ( - 3.43)\]
\[\Large f(12.57)= 4- \frac{3.43}{8} = > \frac{32- 3.43}{8} = 3.57\]
Note : this is a rough approximation.

Answer 17

that earns a metal for hard work

Answer 18

thats kinda the same method