Answer:
1.C
2.A
3.D
4.C
5.B
6.A
7.B
8.C
9.A
10. D
Step-by-step explanation:
For Problem 1, the work is as follows:
For Problem 2, the work is as follows:
For Problem 3, the work is as follows:
For Problem 4, the work is as follows:
For Problem 5, the work is as follows:
For Problem 6, the work is as follows:
For Problem 7, the work is as follows:
For Problem 8, the work is as follows:
For Problem 9, the work is as follows:
For Problem 10, the work is as follows:
Hope this Helps!
Answer:
45
Step-by-step explanation:
Two tangents drawn to a circle from an outside point form arcs and an angle, and this formula shows the relation between the angle and the two arcs.
m<EYL = (1/2)(m(arc)EVL - m(arc)EHL) Eq. 1
The sum of the angle measures of the two arcs is the angle measure of the entire circle, 360 deg.
m(arc)EVL + m(arc)EHL = 360
m(arc)EVL = 360 - m(arc)EHL Eq. 2
We are given this:
m<EYL = (1/3)m(arc)EHL Eq. 3
Substitute equations 2 and 3 into equation 1.
(1/3)m(arc)EHL = (1/2)[(360 - m(arc)EHL) - m(arc)EHL]
Now we have a single unknown, m(arc)EHL, so we solve for it.
2m(arc)EHL = 3[360 - m(arc)EHL - m(arc)EHL]
2m(arc)EHL = 1080 - 6m(arc)EHL
8m(arc)EHL = 1080
m(arc)EHL = 135
Substitute the arc measure just found in Equation 3.
m<EYL = (1/3)m(arc)EHL
m<EYL = (1/3)(135)
m<EYL = 45
Combinatorial Enumeration. That whole class was a rollercoaster ride of mind-blowing generating functions to prove crazy things. The exam had ridiculous questions like 'count the number of cactus trees with n vertices such that etc etc etc' and you'd do three pages of terrible terrible sums and algebra. Then your final answer would be something beautiful like n/2 and you'd breath a sigh of relief and thank the math gods.
After five weeks they will have the same amount of money.
