Question

What is the exact distance from (–4, –2) to (4, 6)?

Answer 1

Use the distance formula
\[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Answer 2

I tried to do it by: 8^2 + 4^2 = c^2

Answer 3

distance formula is derived from pythogorian theorem

Answer 4

64+16 = c^2
80 = c^2

Answer 5

I haven't learned that yet

Answer 6

answers are 24, 88 100, or 128 all sq roots

Answer 7

\[\huge d=\sqrt{(4--4)^2+(6--2)^2}\]
\[\huge d=\sqrt{(4+4)^2+(6+2)^2}\]
\[\huge d=\sqrt{(8)^2+(8)^2}\]
\[\huge d=\sqrt{64+64}\]
\[\huge d=\sqrt{128}\]
\[\huge~\rm~d=\sqrt{128}~or~11.31\]

Answer 8

thanks. how do I give medals? This is my first question.

Answer 9

You may hit the best response button = ) ONLY if you are satisfied with the help.

Answer 10

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Answer 11

You can also simplify \(\Large \sqrt{128}\) like so


\[\Large \sqrt{128} = \sqrt{64*2}\]

\[\Large \sqrt{128} = \sqrt{64}*\sqrt{2}\]

\[\Large \sqrt{128} = 8\sqrt{2}\]

Notice how I factored 128 into 64*2. One of the factors (64) is a perfect square.