Step-by-step explanation:
Divide two on both sides to get rid of it and the make the equation in the form y = mx + c

y = 3/2x + 5
since both lines are parallel they must have the same gradient which is 3/2
y = 3/2x + c
All you have to do now is to replace x and y with (2, -5) to find c
x = 2
y = -5
-5 = 3/2 × 2 + c
-5 = 3 + c
c = -5-3
c = -8

Answer:
54 coins left
Step-by-step explanation:
72 x .25 = 18
72 - 18 = 54
*see attachment for diagram
Answer:
Perimeter = 38
Step-by-step explanation:
Recall: when two tangents are drawn to meet at a point outside a circle, the segments of the two tangents are congruent.
Given,
CQ = 5
PQ = 10
PR = 14
Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR
CQ = QB = 5 (tangents drawn from an external point)
BP = PQ - QB
BP = 10 - 5 = 5
BP = PA = 5 (tangents drawn from an external point)
AR = PR - PA
AR = 14 - 5 = 9
AR = RC = 9 (tangents drawn from an external point)
✔️Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR
= 9 + 5 + 5 + 5 + 5 + 9
Perimeter = 38
3.
4^3 is equal to 64. Therefor log4(64) must be equal to 3.