PLEASEEE HELP FAST 2 QUESTIONS

k = 12 for the first problem. p = 50 AND q = 80 for the second problem.

Step-by-step explanation:

Problem 1

I will call the center point of the first problem where the two lines intersect E. Angle AEC is an angle we need to know to solve this.

We can find it by subtracting 174 from 180, since it forms a straight like with Angle AED.

So, Angle AEC is 6 degrees.

Notice how the right angle on the bottom, the angle labelled 4k, the angle labelled 3k, and Angle AEC are all on one side of Line CD. This means that added up, they all have to equal 180 degrees.

Let's make an equation since we know the values of all these angles.

The right angle is given to be 90 degrees and we found that Angle AEC is 6 degrees. The other two angles are 4k and 3k.

180 = 90 + 4k + 3k + 6

Now let's solve for k

180 = 96 + 7k

84 = 7k

k = 12

Problem 2

Once again, I will call the center point where the two lines intersect, E.

Angle AEC is given to equal 2p and Angle BED is given to be (p + 50).

They are vertical angles and we know they are both equal, so we can set up an equation to find p

2p = p + 50

Solve for p using basic algebra.

p = 50

Now, we don't even need to solve for q because there is only one option that says p = 50, the third one.

3x^3-x^2+6x-9

2,600 divided by 80 = 32.5

step-by-step explanation: