C is the incenter of isosceles triangle ABD with vertex angle ∠ABD. Does the following proof correctly justify that triangles ABC and DBC are congruent? It is given that C is the incenter of triangle ABD, so segment BC is an altitude of angle ABD. Angles ABC and DBC are congruent according to the definition of an angle bisector. Segments AB and DB are congruent by the definition of an isosceles triangle. Triangles ABC and DBC share side BC, so it is congruent to itself by the reflexive property. By the SAS postulate, triangles ABC and DBC are congruent. Triangle ABD with segments BC, DC, and AC drawn from each vertex and meeting at point C inside triangle ABD. There is an error in line 1; segment BC should be an angle bisector. The proof is correct. There is an error in line 3; segments AB and BC are congruent. There is an error in line 5; the ASA Postulate should be used.

egaragag

Step-by-step explanation:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)