Triangle \triangle A'B'C'△A ′
B
′
C
′
triangle, A, prime, B, prime, C, prime is the result of dilating \triangle ABC△ABCtriangle, A, B, C about point BBB by a scale factor of \dfrac{4}{3}
3
4

start fraction, 4, divided by, 3, end fraction.

Determine whether each claim about the properties of \triangle ABC△ABCtriangle, A, B, C and \triangle A'B'C'△A
′
B
′
C
′
triangle, A, prime, B, prime, C, prime is true or false.
The lengths of \overline{AB}
AB
start overline, A, B, end overline and \overline{A'B'}
A
′
B
′

start overline, A, prime, B, prime, end overline are equal.

The areas of \triangle ABC△ABCtriangle, A, B, C and \triangle A'B'C'△A
′
B
′
C
′
triangle, A, prime, B, prime, C, prime are the same.

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

17   x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation:

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )