<p>Part II (26-50):</p>
<ol>
<li><p>D: 1000 is the initial amount, so P=1000. A is the final amount, so A=5000. Plug into the equation: 5000=1000e^.08t Solve to get t=20.1</p></li>
<li><p>B: If sin(x)>0, then x must be in quadrants I or II, where sin (y) is positive. If sinx is positive, then cosx must be negative in order to satisfy the inequality. cos (x) is negative only in quadrants II and II, so x can only be in quadrant II to satisfy all conditions.</p></li>
<li><p>C: For every x and ?x in the function, there is an equal f(x) for both values. This means that 3 and -3 creates the same value. f(3)=f(-3) f(3)=8, so f(-3)=8, and so (-3,8)</p></li>
<li><p>E: If x^2=y^2, then x can be equal to y, but x and y can also be negatives of each other. For instance 25=25. x can be 5 or -5, and same for y. E is not true. </p></li>
<li><p>D: 9 possible students in first position, 8 in next, 7, then 6, etc. This is 9!, which is 362880.</p></li>
<li><p>Graph and use table function, starting at 1 and going by small intervals of say .1. You can see that from either side, the function approaches 1 as x approaches one, though it is undefined at x=1.</p></li>
<li><p>C: f(2)=1. Graph f(x), and trace all the values until one of them equals 1. f(4/3) gives 1.</p></li>
<li><p>Graph this, with window ranges of x in the answer choice. I used this method to get the answer while taking the test, but it took forever. But the back of the book says something like this: the period of y is the same as the period of tan(3(pi)x), since only the coefficient inside the trig affects the period. The period of tan(x) is pi, and the coefficient of x in y is 3pi, making the period shrink. Since its 3pi coefficient, then multiply 1/3pi by pi, which give 1/3. </p></li>
<li><p>C: Draw a right triangle. The hypotenuse is 50, and the East Road is 20 miles. Use the Pythagorean Theorem to find the North Road, which is equal to 45.83. Add up the north and east roads, which is 65.83. Distance divided by velocity gives time, which is 1.46 hours. .46 hours is 27.6 minutes, so it takes about 1 hour and 28 minutes.</p></li>
<li><p>C: f(x) equals 0 at -1 and 2, so -1 and 2 must be zeros of the function f(x). (x+1) and (x-2) is in f(x). Only choice C has those factors.</p></li>
<li><p>D: A factor of a number must be a multiple of its prime numbers. You can get 10 by 5<em>2. You can get 20 by 2</em>2*5, etc. You cannot get 30 by any multiplication of the prime factors, so 30 cannot be a factor of n.</p></li>
<li><p>E: Graph sin(x) and 3cos(x) in a viewing window of [0,pi/2]. Their intersection gives x. You can also rewrite the equation to tan(x)=3 by dividing cos(x), and then solving for x.</p></li>
<li><p>To find f(x) inverse, switch y and x, and solve for y. f(x) inverse=x^2/50. Plug 10 into this inverse function. f(10)=10^2/50= 2</p></li>
<li><p>C: Defined as the previous two terms added together, so 1,1,2,3,5,8,13,21,34,55?10th term is 55.</p></li>
<li><p>D: Graph the function. It is increasing for x equal to or greater than 3. The function crosses the x-axis at 3 points, so it must have 3 real answers. I is true. It cannot have 2 more nonreal answers since there are already 3 answers to this third power polynomial. II is not true. f(x) is greater than -16 for all xgreater to or equal to zero, since -16 is a local minimum. III is true.</p></li>
<li><p>A: fg(0) has to be 0 since f(0)g(0) is 0. The whole function has to be negative, since f is negative on the left, and g is negative on the right. The only graph that fits all of these conditions is A.</p></li>
<li><p>B: graph both of the functions. The graph continually intersects only at negative numbers and 0. Another method is to see that the x must be negative, because that would make the right side positive. A square root of a number must be positive, and the value inside must positive, and negative squared makes a positive under the radical. Only negative and zero of x works, so B is the answer.</p></li>
<li><p>C: Use law of sine. sin(30)/3=sina(a)/4, 4sin(30)=3sin(a), sin(a)=4sin(30)/3, take inverse sin, a=41.81, sin(41.81)=2/3</p></li>
<li><p>A: The longest segment is the line between two opposite corner vertices. To find this segment length, use the ?Super Pythagorean Theorem?. 8^2+4^2+1^2=length^2 length=sqr.(80), which simplifies to 4sqr.(5)</p></li>
<li><p>A: I don?t know how to get the base into MWord, so just assume the base of a is there! Anyways, log45=log9*5=log9+log5=log3^2+log5=2log3+log5. Since log3=x and log5=y, 2log3+log5=2x=y</p></li>
<li><p>B: Pick any random number for 0<x<pi/2, say pi/3. sin(pi/3)=.86603=t. tan(pi/3)=1.73205 Plug in t=.86603 into each equation to find which one equals to tan(x)=tan(.86603)=1.73205. Only choice B works out.</p></li>
<li><p>E: Graph y=x^2-2x+k, using k as ?a constant greater than 2?, say 4. When you graph this, notice the graph of x^2 shifted to the right 1 unit. You can eliminate choices A, B, and D. The graph shifted 3 units up, which is one less than k where k=4. That means it shifted k-1 up. Right 1 unit and up k-1 units, which is answer choice E.</p></li>
<li><p>A: v=1/3(pi)r^2h This is the volume of a cylinder. .85v=1/3(pi)(kr)^2.92 this is the volume after making the changes, .85(1/3(pi)r^r(h))=1/3(pi)k^2r^2.92 replace v and simplify. Solve for k, which is .96. If you multiply r by .96, you are reducing it by 4%. </p></li>
<li><p>Never learned matrixes, so I won?t even touch this problem!</p></li>
<li><p>A: w is (-a, bi) ?iw=-i(-a, bi)=(ai, -bi^2)=(ai, -b-1)=(ai, b). This means that b is now the real number, and a is the imaginary number. Both are positive, so ?iw must be in the first quadrant. Only A is in the first quadrant.</p></li>
</ol>