Question

Help With Approximation of Irrational's!!

Answer 1

@acxbox22 @iGreen @Haseeb96 @BloomLocke367 @kidrah69 @misty1212 @tylermcmullen23

Answer 2

compare with \(\sqrt {64}. \sqrt{81}\)

Answer 3

ok...

Answer 4

think in terms of what perfect squares are above and below 76.

like as a simple example \[\sqrt{9} < \sqrt{12} < \sqrt{16} \\ 3< \sqrt{12} < 4\]

Answer 5

I think that it is B!

Answer 6

why?

Answer 7

uhhh....there ra mind....thats not it.....

Answer 8

So we can eliminate B and D....

Answer 9

right?

Answer 10

Think about @Empty's working, can you try to do something similar?

Answer 11

I Dont Understand it....

Answer 12

do you know what 8*8 = 8^2 = ?

Answer 13

yes

Answer 14

what is it equal to?

Answer 15

64

Answer 16

good, so \[64=8^2\]
this means that
\[\sqrt{64} =8 \]

Answer 17

yes...

Answer 18

do you know the square root of forty-nine?
\[\sqrt{49}\]

Answer 19

7

Answer 20

ohhh...I see it's A!

Answer 21

\[\sqrt {49}<\sqrt {64}\]because\[7<8\]

But what about \(\sqrt{76}\)? could that be between 7 and 8?

Answer 22

Do you know \[\sqrt{81}\]?

Answer 23

9

Answer 24

Was it B, this whole time?

Answer 25

\[\sqrt {49}<\sqrt {64}<\sqrt{81}\]because\[7<8<9\]

where does \[\sqrt {76}\] fit in?

Answer 26

8-9

Answer 27

good.

Answer 28

Make sense now?

Answer 29

I said that at the Beginning!!!!

Answer 30

Oh well, at least i know how to solve it know...

Answer 31

will you give me a medal and be my fan?

Answer 32

Yeah, you weren't convinced before,

If you check on a calculator
\[\sqrt{76} = \texttt{8.717797887...}\]