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nirvana33 [79]
2 years ago
5

Triangle \triangle A'B'C'△A

Mathematics
1 answer:
Harrizon [31]2 years ago
6 0

From the information given, the rotated triangle is same as the original triangle. This is because, vertices of the rotation remain at the origin - (0,0).

<h3>What is the proof the above?</h3>

Notice that the question states:  △A'B'C' is the image of △ABC. Given that the △ without primes is the original and the one with primes depict the rotated triangle,

under a rotation about the origin, (0,0) (0,0) (0, 0), indicating no movement,

△A'B'C' is in the same position with △ABC.

Learn more about rotation at:
brainly.com/question/27915888
#SPJ1

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Graph the line with slope -1/2 passing through the point (5,4)
slava [35]

Answer:

put a down a point at (5,4) and go down 1 right 2, then plot another point.

You also can go from (5,4) up 1, left 2 and plot a point.

6 0
2 years ago
Algebra 1 , PLEASE PLEASE help . and if you know how to do algebra i'll have more question you can answer .. if you want :, )
Ostrovityanka [42]

Answer:

It's A, Because it makes the most sense in this algebra question.

Step-by-step explanation:

Have a great day!

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8 0
3 years ago
Read 2 more answers
2
Aleksandr [31]

Answer: lower bound = 9550

<u>Step-by-step explanation:</u>

Since the number is rounded to the nearest hundred, the actual value is somewhere between 9550 (which rounds up to 9600) and 9649 (which rounds down to 9600).

The lower bound is: 9550

The upper bound is: 9649

6 0
3 years ago
I need to know this asap!
Vinvika [58]

w+2x=16 ⇒ w=16-2x


x+2w=14        x+2(16-2x)=14

                      x+32-4x=14

                     -3x=14-32

                      -3x=-18  /:(-3)

                       x=6


w=16-2x=16-2·6=16-12=4


2y+w=-6 ⇒   2y+4=-6

                   2y=-6-4

                   2y=-10  /:2

                   y=-5

4 0
3 years ago
Find the distance between the two points in simplest radical form.<br> (-1,9) and (4, -3)
jolli1 [7]

Answer:

GIVEN BY POINTS :-

( -1, 9) , ( 4 , -3)

Here ,

\bullet  \: \:  x _{1} = - 1 \\ \bullet \: \:  x _{2} = 4 \\ \bullet \: \:   y _{1} = 9 \\ \bullet \: \:    y _{2} =  - 3 \\

Using Distance formula :

=  >  \sf d = \sqrt{( {x _{2} - x _{1}) }^{2} + {( {y _{2} - y _{1}) }^{2} } } \\ \\  =  > \sf  d = \sqrt{ ({ 4  + 1)}^{2} + { ( - 3 - 9)}^{2} } \\  \\ =  >  \sf d = \sqrt{ {5}^{2} + {( - 12)}^{2} }  \\ \\  =  > \sf d = \sqrt{25 + 144}  \\ \\  =  > \sf d = \sqrt{169}  \\ \\ =  > \boxed{ \sf{  d = 13\:units }}\\

8 0
3 years ago
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