Answer:
Step-by-step explanation:
we know that
If line I s a perpendicular bisector of line segment R Q at point T
then
T is the midpoint of line segment RQ
and
we have
so
Find the value of x
we have
substitute the value of RT
solve for x
Find the value of SQ
substitute the value of x
we have that
Triangle RTS and Triangle QTS are congruent by SAS
see the attached figure to better understand the problem
so
therefore
Answer: 9a-9b-6
7(3a−2b)+5b−3(4a+2)
Distribute:
7(3a−2b)+5b−3(4a+2)
=(7)(3a)+(7)(−2b)+5b+(−3)(4a)+(−3)(2)
=21a+−14b+5b+−12a+−6
Combine Like Terms:
=21a+−14b+5b+−12a+−6
=(21a+−12a)+(−14b+5b)+(−6)
=9a+−9b+−6
Answer:
y=.15x+45
$120
Step-by-step explanation:
.15*500=75
75+45=120