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.

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Step-by-step explanation:

Just took the test

A edg 2020

Step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

50.6

step-by-step explanation: