Solution #16

(5x - 11)/(2x²+x-6) = (5x-11)/((2x - 3)(x+2)).

We then write (5x-11)/((2x-3)(x+2)) = A/(2x-3) + B/(x+2).

Multiplying both sides by (2x-3)(x+2), we have:

5x - 11 = A(x+2) + B(2x-3)

5x - 11 = Ax + 2A + 2Bx - 3B

5x - 11 = (A+2B)x + (2A-3B)

Equating coefficients of like terms, we have:

A + 2B = 5
2A - 3B = -11

Doubling the first equation gives 2A + 4B = 10, and subtracting the second from the first gives 7B = 21, and B = 3. Then A = -1, so our sum is:

-1/(2x-3) + 3/(x+2)

Problem #18

[This problem is worth 8 points.]

Find (Σ_n=1⁴⁴cos n^°)/(Σ_n=1⁴⁴sin n^°)

Problem #17

[This problem is worth 1 point.]

A car travels from point A to point B at a speed of 2 miles per minute. It then returns from point B to point A at a speed of 2 minutes per mile. What is its average speed for the entire journey, in miles per hour?

Problem #16

[This problem is worth 2 points.]

Express the fraction (5x - 11)/(2x²+x-6) as the sum of fractions each of which has as its denominator a polynomial of degree 1.

Solution #15

First note that triangles BCM and BPM are both right, because CM is chosen perpendicular to BK. Because BK is the angle bisector of angle B, BCM and BPM are similar. Because they share side BM, they are congruent.

Thus M is the midpoint of CP, and BC = MP = 120.

Similarly, triangles AQN and ACN are congruent. Thus N is the midpoint of CQ, and AQ = AC = 117.

Because M and N are respectively the midpoints of PC and QC, by considering triangle PCQ we see that MN is half of PQ.

AQ + BP = AB + PQ, so PQ = AQ + BP - AB = 120 + 117 - 125 = 112. Thus MN = 56.

Week 3 Standings

Through problem #15:

SS: 54
MK: 46
AG: 29
JP: 7
BW: 7
ML: 6
TM: 5
PG: 4
SD: 3
SG: 3
MD: 1

Solution #14

Recall that, using the Binomial Theorem, the expansion of (x+y)ⁿ is the sum of all terms of the form nCk*x^n-ky^k.

We want a term in the expansion of  (a-1/ √ a  )⁷ in which the a term has an exponent of -1/2. Using the above general form, the k^th term will have an exponent of 7-k (from the a term) plus (-1/2)k (from the 1/ √ a  term). So we need 7 - k - k/2 = -1/2, or 15/2 = 3k/2, or k = 5. The coefficient is then (-1)⁵*7C5 = -21.