January 2009 Algebra II Individual Answer Sheet

1.) -2(x+

-2(x-)

= - D

2.)Evaluate (1-i)

((1-i)) = (-2i)= 8i

3.) -2x+ (-3 x) - 4(x+ 2x)

-2x+ -27 x - 4x- 8x

= -27x-6x-8x

4.)

Test Boundary Points on a Number Line with a closed circle at 3 and open circles at 2 and -2.

5.) . Multiply top equation by x+1 and 2nd equation by x-1. The x-1 in the denominators cancel leaving the equation as C.

6.) The parabola opens to the left and has a vertex of (0,-6)so the line of symmetry would be a horizontal line at y= -6. D

7.)Find the sum of the slope, x-intercept, and y-intercept for a line passing through the points (-1,5) and (2,-4).

Slope = -3 , use y=3x+b and plug in a point to find b. B= 2 and then you get y=3x+2 plug in a 0 for y to find the x-intercept of 2/3.

-3 +2 +2/3= -1/3 B

8.)Find the sum of the vertical asymptote(s) for g(x) if g(x)=.

Vertical asymptote is at x=3. There is hole at x=2

9.)Solve the absolute value equation . Solve the equation as is 1/2x+2= 3/4x -2. x= 16. Then create one side to be negative -1/2x-2= 3/4x+2. x=0

10.) Find the area of the polynomial region defined by the solution set to the system below.

2y-x

2y+x

The points of intersection that make up the shaded region are (0,0), (0,12), (0,5), (12,5), and (6,8). The figure forms a rectangle and a small triangle. The rectangle is 5 x 12= 60 and the triangle is (3x12)/2= 18. 60 + 18= 78. C

11.) Solve the exponential equation by expressing the solution in terms of logarithms. Rewrite as . E

12.) 130/5= 26 130/25= 5 130/125= 1

26+5+1= 32 D. Note: Remember than 2*5 gives you zeros. There are plenty of 2’s so find how many 5’s there are (32 of them). D

13.) (8+.05x)(600-6x)= -.3x-18x+4800. Find the Vertex -b/2a. 18/-.6= -30= x. Plug back in your cost equation $8+ (.05)(-30)= $6.50. B

14.) = n= 4+ = n-4n-12=0

(n-6)(n+2)=0 n=6 (solution has to be positive).

15.) If n is an odd integer, then each of n − 1 and n + 1 is even.

In fact, n − 1 and n + 1 are consecutive even integers, so one is a multiple of 4 and the other is divisible is 2 (since it is even).

Thus, (n − 1)(n + 1) contains at least 3 factors of 2, which tells us that (n − 1)(n)(n + 1) does as well, ie. is divisible by 8. So if n is an odd integer, then(n − 1)(n)(n + 1)/8 is an integer.(There are 39 odd integers between 2 and 80, inclusive.)

If n is an even integer, then each of n − 1 and n + 1 is odd.

Thus, (n − 1)(n)(n + 1) is divisible by 8 only when n is divisible by 8.

(There are 10 multiples of 8 between 2 and 80, inclusive.)

Therefore, there are 39 + 10 = 49 integers.

16.) Use harmonic mean = 5mph A

17.) Divide by x-2i. Use remainder theorem, (2i)-2(2i)+3(2i) -6= 16 +16i +6i-6= 10 +22i . D

18.)For a given cubic polynomial in the formthe sum of the roots taken two at a time is . So ¼ E.

19.)Complete the Square to get (x-8x +16) + 4(y+y+1/4)= 12+16+1

Formula for the area of an ellipse is ab =29/2 B

20.) (1+1+1+1)= =2 D

21.) = 10= 91 digits E

22.)If x+y=30 and (x+y)=56. Find (x-y).

x+2xy+y=56 substitute 30+2xy=56 2xy=26 .

x-2xy+y= substitute 30-26= 4 C

23.) 8= 12800 D

24.) Find the constant term in the expansion of .

12 C 9 (x)(= -1760 D

25.)1001 + 1111+ 1221 + 1331 +1441 + 1551= 7656 A

26.)Identify the reflection of the graph of y= (x-3)-5 over the line x = -8.

The vertex is (3, -5) and the reflection shifts the graph left 11 more than -8. So the new vertex is (-19,5) and the graph still faces up. D

27.)Find the units digit in the expansion of .

3= follows the pattern 3, 9,7, 1 and so ends in 1.

2= follows the pattern 2,4,8,6 and so ends in 8.

6= follows the pattern 6…. and so 6 to any positive power ends in 6.

So 1*8*6= 48 so the answer is A.

28.) So

Cross-multiplying, we obtain 11(b − 2b) = 6a.

Since 11 is a divisor of the left side, then 11 must be a divisor of the right side, that is, a must be divisible by 11.Thus, let a = 11A, with A as a positive integer. So we get 11(b − 2b) = 6(11A) or b − 2b = 6A. Since 6 is a divisor of the right side, then 6 must be a divisor of the left side. What is the smallest positive integer b for which 6 is a divisor of b2 −2b? We can quickly check that if b equals 1, 2, 3, 4, or 5, then b − 2b is not divisible by 6, but if b = 6, then b − 2b is divisible by 6.Therefore, for the smallest values of a and b, we must have b = 6, so 6A = 62 − 2(6) = 24,whence A = 4 and so a = 11A = 44.Thus, the smallest possible value of a + b is 44 + 6 = 50. D

29.) If x varies jointly with y and z, and x=96 when y=4 and z=2. Find z when x=108 and q=3. so K=3. and z=4. A

30.) = 27 So x=27 (-4)+ 27 = 19 A