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gizmo_the_mogwai [7]
3 years ago
9

Each leg of a 45°-45°-90° triangle measures 14 cm. What is the length of the hypotenuse?

Mathematics
2 answers:
4vir4ik [10]3 years ago
6 0

For a 45-45-90 angle the hypotenuse is s * square root of 2

so the hypotenuse is 14 * square root of 2 or 19.8

:)))

poizon [28]3 years ago
3 0

Answer:

Its D) 14 squareroot 2

Step-by-step explanation:

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Any fraction

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Rotate the figure 90 degrees about vertex A.
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A' is (1,1) B is (4,1) C is (1,-1)

Step-by-step explanation:

Since we rotating the figure about point a, we know a is the center of the rotation meaning no matter how far we rotate point a new image will stay on where point a pre image was which in this case is (1,1). Also since we know the rules of rotating a angle 90 degrees About the origin we are going to translate the figure to have the one point we are rotating about at the orgin. Since translations are a rigid transformations, the figure will stay the same A. Move the figure 1 to the left and 1 down so A becomes 0,0 B becomes 0,3 and C becomes 2,0. Then apply the rules of 90 degree clockwise rotation rules. (x,y) goes to (y,-x) . A stays (0,0) B becomes (3,0) and C becomes (0,-2). Then translate the figure 1 to the right and 1 down since we rotating about point a which is 1,1 and it at 0,0 rn. A' is 1,1. B' becomes (4,1). C' becomes (1,-1).

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Divide and round to the nearest hundredths place.<br> 4.05 ÷ 0.07
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Find the distance between the endpoints (7,8) (4,2) A.10.8 B.2.2 C.6.7 D.14.9
Paha777 [63]

Answer:

\boxed{  \bold{ \huge{ \boxed{ \sf{6.7}}}}}

Option C is the correct option.

Step-by-step explanation:

Let the points be A and B

Let A ( 7 , 8 ) be ( x₁ , y₁ ) and B ( 4 , 2 ) be ( x₂ , y₂ )

<u>Finding</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>between</u><u> </u><u>these</u><u> </u><u>points</u>

\boxed{ \sf{distance \:  =  \:  \sqrt{ {(x2 - x1)}^{2} +  {(y2 - y1)}^{2}  } }}

\dashrightarrow{ \sf{ \sqrt{ {(4 - 7)}^{2}  +  {(2 - 8)}^{2} } }}

\dashrightarrow{ \sf{ \sqrt{ {(  - 3)}^{2} +  { (- 6)}^{2}  } }}

\dashrightarrow{ \sf{ \sqrt{9 + 36}}}

\dashrightarrow {\sf{ \sqrt{45} }}

\dashrightarrow{ \sf{6.7 \:  \: units}}

Hope I helped!

Best regards! :D

6 0
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