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patriot [66]
3 years ago
12

Complete the paragraph proof. Given: and are right angles Line segment A B is-congruent-to line segment B C Line segment B C is-

congruent-to line segment A C Prove: Line A R bisects Angle B A C Triangles A B R and R C A share side R A. A line is drawn from point B to point C and intersects side A R at point P. It is given that and are right angles, and . Since they contain right angles, ΔABR and ΔACR are right triangles. The right triangles share hypotenuse , and reflexive property justifies that . Since and , the transitive property justifies . Now, the hypotenuse and leg of right ΔABR is congruent to the hypotenuse and the leg of right ΔACR, so by the HL congruence postulate. Therefore, ________ by CPCTC, and bisects by the definition of bisector.
Mathematics
2 answers:
aleksklad [387]3 years ago
7 0

Answer:

<BAR ≅<CAR

Step-by-step explanation:

Just took the test

anygoal [31]3 years ago
5 0

Answer:

A edg 2020

Step-by-step explanation:

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15ft because 12x.25=3 so 3ft is 25% of the 12ft, add the 3+12=15
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What number is 24% of 80%
artcher [175]
The answer is 16 hope this helps you.
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The hypotenuse of a right triangle measures 6cm and one of its legs measures 3cm.Find the measure of the other leg. If necessary
belka [17]

Answer:

5.2

Step-by-step explanation:

a^2+b^2=c^2

Where c is the length of hypotenuse

and a, b are lengths of the legs.

Put in 6 for c,

Put in 3 for a and you get 9+b^2=36

Subtract 9 from both sides and get b^2= 27

Then find the square root of both sides to get b, so the answer is the square root of 27.

The cloest square root is 5, then apporixmate the tenths value and see if they work.

Answer: 5.2

8 0
2 years ago
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What is the image of the point (6, -4) when rotated 180 degrees anticlockwise about the origin?
icang [17]

Answer:

What is the image of the point (6, -4) when rotated 180 degrees anticlockwise about the origin?

(-6, -4)

(-6, 4)

(4, 6)

(-4, -6)

the answer is (-6, 4).

8 0
3 years ago
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Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck
Nimfa-mama [501]
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}
2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C


4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C


y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}

At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}

4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}


4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C
0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C
C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
6 0
3 years ago
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