Answer:
8b-5=6b+23
8b-6b=23+5
2b=28
b=14
Step-by-step explanation:
Answer:
Part A
x is 46°
Part B
Alternate angles are angles that are in relatively opposite locations relative to a transversal
Please see attached diagram showing alternate angles
Step-by-step explanation:
Part A
∠DRP = 110° (Given)
∠QPA = 64° (Given)
∠QPR =
Given that AB is parallel to CD, we have;
∠DRP is congruent to ∠APR (Alternate angles to a transversal RP of parallel lines AB and CD)
Therefore, ∠APR = 115°
∠APR = ∠QPA + ∠QPR (Angle addition postulate)
∴ 115° = 64° + ∠QPR
∠QPR = 110° - 64° = 46°
x = 46°
Part B.
Given that AB is parallel to CD, the lines common (that intersects) both lines are the transversal lines
The angles formed between the parallel lines and the transversal lines have special relationships based on their position with respect to each other
In the question, the angle 110° given between CD and the transversal RP, is found to at an alternate position to the angle ∠APR between the same transversal RP and AB and given that alternate angles are always equal, angle ∠APR is therefore also equal to 110°.
Answer:
A' (1, 2)
Step-by-step explanation:
You can tell so from the graph. Since it's A' and not A, you don't have to rotate the triangle in order to find the solution.
Step One
Find the complimentary angle of 28o Compliments add to 90o.
28 + c = 90
c = 90 - 28
c = 62o
Step two
Find the sin of 62o
sin(62) = 0.8829
So the answer is B
Another way to do it is to take the cos of 28
Cos(28) = 0.8829 which is still B.
Answer:
120 in
Step-by-step explanation:
The area of a parallelogram is its base multiplied by its height.

For this problem, the base is 15 in while the height is 8 in. Plug in the numbers into the equation and solve.



The area of the parallelogram is 120 in.