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melstutes
 one year ago
Suppose that the endpoints of one leg of a 45°45°90° triangle are (0, 3) and (4, –1). What is the length of the hypotenuse?
melstutes
 one year ago
Suppose that the endpoints of one leg of a 45°45°90° triangle are (0, 3) and (4, –1). What is the length of the hypotenuse?

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melstutes
 one year ago
Best ResponseYou've already chosen the best response.04 – 0 = 4 1 3= 4 4 sq + 4 sq = c sq 16 + 16 = c sq √32 = c The distance between the two points is √(16+16) = √32 So, the hypotenuse is √32*√2 = √64 = 8 is this right?

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0I thought distance was same as hypotenuse. Why do you have to multiply by sq root of 2

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0This was also given. Hint: Remember how the sides relate in a 45°45°90° triangle.

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0I kept getting √32 as my answer.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2using theformula of distance we can write: \[d = \sqrt {{{\left( {4  0} \right)}^2} + {{\left( {  1  3} \right)}^2}} = \sqrt {16 + 16} = \sqrt {32} = 4\sqrt 2 \]

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0Ok  I did that and that is what I thought was the hypotenuse

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! the length of the hypotenuse is \[\sqrt {32} = 4\sqrt 2 \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so the length of each side is: x=4

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0That is not a choice. Answers include 8 units, 4 units, 10 units or 2 units. I was told to multiply by 2 and answer was 8 but without an explanation. I don't understand why

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I think that your exercise asks for the value of x. So if \[x\sqrt 2 = 32 = 4\sqrt 2 \] then we get: \[x = 4\]

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0Question is what is length of hypotenuse so I am lost

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops..I have made an error, hereare the right steps: \[\begin{gathered} x\sqrt 2 = \sqrt {32} = 4\sqrt 2 \hfill \\ x = 4 \hfill \\ \end{gathered} \] yes, I know, nevertheless if I look at your drawing: dw:1434476827731:dw I understand that we have to determine the value of x

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2namely in your exercise there are 2 parts: the first one asks for the hypotenuse, using the coordinates which you have provided the second part asks for the value of x, using the length of the hypotenuse computed in the first part

melstutes
 one year ago
Best ResponseYou've already chosen the best response.0Oh, ok. Question is misleading or confusing.
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