Math 120. x x 4 x. . In this problem, we are combining fractions. To do this, we must have

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1 Math 10 Final Eam Review Worked out solutions. In this problem, we are subtracting one polynomial from another. When adding or subtracting polynomials, we combine like terms. Remember that like terms are those terms having the same variable to the same power. Thus we have = In this problem, we are combining fractions. To do this, we must have a common denominator. Remember that the least common denominator (LCD) is the simplest epression that is a multiple of both denominators. To find the LCD, we need to factor the denominators. We see that and Thus the LCD is So we convert the original epression: Notice that we had to multiply the numerator and denominator of the first fraction by 1, since this was the factor missing in the denominator. Similarly, the second fraction was missing a factor of 4 in the denominator, so we multiplied above and below by 4. Now that we have a common denominator, we combine the fractions In general, we need to reduce the fraction, if possible. But in this case, we see that the numerator is not factorable, so the fraction here is not reducible. So we re done.. This question asks if (, ) is a solution of the system 4 y 14. 4y 14 Remember that a solution is something that makes a statement true. In this case, the statement is a system of two equations, so it is a statement comprised of two statements. Thus in order for the system to be true, both equations must be true. So we re looking to see if (, ) satisfies both equations. So we plug in for and in for y in both equations: 4() + = 14 (true) () + 4() = 14 (true) We use long division The remainder is 6, so we see the answer is

2 a b 4 a a b b 4a b a b. 8 5 t 4 is a sum of cubes. t 4 t 7 t 7t 7t This is a difference of squares We are looking for two numbers whose product is 15(6) and whose sum is 19. The two numbers are 10 and 9. This enables us to break up the linear () term into two terms so we can group the polynomial and factor y. We see that the first three terms make up a perfect square trinomial, so the entire epression is a difference of squares y y y y y. 10. To find the equation of the line through 5, and perpendicular to the line 5 y 7, we need to know the slope of the line (we already know a point on the line). 5 The slope of the line 5 y 7 is. We know that the slope of our line is the negative reciprocal of this ie..,, since the two lines are perpendicular. So the equation of 5 the line is y b 5 where b is the y-intercept. We know that the line passes through (-5, -), so this point satisfies the equation. Thus 5 b 5 b b 0 Thus we have y. 5 fg f 4 7 is not one-to-one. For eample, f f , so a singe y value corresponds to two different values. (You could also say that the graph fails the horizontal line test). f 5 is one-to-one, and to find the equation of the inverse function, we swap 1. and y in the original equation and solve for y. y 5 5y 5 y 5 1 y f k 10k 16 k 5k k 8k k k 5 k 14. k 1k 40 k k 8 k 8 k 5 k 4 k k 4

3 loga m loga n loga j loga k a m a n a j a k log log log log (power property of logarithms) log log a m n 1 4 m j 5 a 5 nk j k (product property of logs) (Note that 6 value sign). 4 8 (note that since 6 cannot be negative, but could be, so we need the absolute cannot be negative, we don t need the absolute value sign) a y. We substitute 5a in for y in the first equation to get y 5a 7a 5a 7a 10a a 10, y 5a a a 0. 8 f 9 7. An absolute value can never be negative, so this equation is never true, no matter what the value of f is. Thus there is no solution (the solution set is the empty set) We re looking for the solution set. Remember that a solution for any statement is something (in this case a value of ) that makes the statement true. Well, we know that an absolute value epression can never be negative, so the epression on the left is always greater than 9, no matter what is. Thus the inequality is true for all values of (i.e., all real numbers) We can clear fractions by multiplying both sides of the equation by the LCD, which is 0. When we do this, we get y y The easiest way to solve this is to substitute for y 1. Then we have

4 or 5 4 Now since = y 1, we see that y = + 1, so y or 4. log 4 log 4 log. We want to use the same principle we used in problem But to do this, we need to epress the left side as a single logarithm. We use the quotient property of logarithms to do this: 4 log log log log Both sides of the equation are base- logarithms. Since all log functions are one-to-one, we know that if the base- log of one thing equals the base- log of another thing, then the two things must be the same. In other words, we can just equate the arguments of the logarithms to get or We use a similar strategy as that in problem 6. 1 log 5 log 7 log 5 1 log 7 log 5 log 7 log 7 log 5 log 7 log 7 log 5 log 7 log 7 log log 5 log Notice that the variable we re trying to solve for is in the eponent. To change this, we need to take a logarithm of both sides and then use the power property of logs.

5 5 1 ln ln ln ln 5 ln ln ln ln ln 5 ln e. Since we see that is in the eponent of the natural eponential function, we take the natural log of both sides to remedy this situtation ln e ln 0. ln ln Let s define to be the thing we re asked to find. That is Let = the number of liters of 10% alcohol solution needed. Now we write an equation that states that the amount of alcohol we start with is equal to the amount of alcohol we end up with. 10% 70 70% % (70) Remember what it means to have three consecutive odd integers. An eample of three consecutive odd integers is 7, 9, 11. So let s define to be the first one. Then + is the second one, and + 4 is the third. Then the first sentence in the word problem translates to the equation Thus the third integer is For compound interest, the formula is AP 1 r n nt

6 where A is the amount after t years, P is the original principal invested, r is the annual interest rate, and n is the number of compoundings per year. Since we re compounding quarterly, n = 4. Let s plug in all the given values and solve for the time that corresponds to an amount of $15, t 4t 4t ln 5 ln 1 4t ln ln 5 t. years 4ln 1. We see that the original amount (i.e., the amount when t = 0) is 400. Thus we want to know how long it will take for the amount to be 00. So here goes... t A t 400e 00 e t t ln 1 1 t ln.9 years We know the formula for continuously compounded interest is rt A Pe where A is the amount after t years, P is the initial amount invested, and r is annual interest rate. We are asked to find the time that corresponds to an amount of $,000, so we solve for t: rt A Pe A rt e P A ln rt P 1 A t ln ln 7.7 years r P Since we re asked to find the number of nickels, let s define accordingly. Let = number of nickels Then 9- = number of quraters Now we write an equation that deals with the value of the coins. Since each nickel is worth $.05 and each quarter is worth $0.5, we have $.05 $.5 9 $

7 Thus there are 0 nickels. 5. Perhaps the easiest way to solve this problem is to put the equation in verte form. h t 15t 90t 15 t 6t 15t 6t9 159 So we see that the maimum height is 15 feet. 15 t Remember that for any quadratic function f a b c, the -coordinate of the verte is given by In this problem, we see that b h. a h t 10t 00t is a quadratic function, and since the lead coefficient is negative, the graph will be a parabola that opens downward. So the maimum height will occur at the verte. Thus the time corresponding to the maimum height is 00 t 15 seconds Since we re asked to find how long it would take Juanita working alone, let s define to mean that. Let = time it takes Juanita to complete the job alone. Now we use the fact that work = rate(time) to get an equation. We know that the time Rick 1 job and Juanita work together is 6 hours. Juanita s rate is, and Rick s rate is hours 1 job. The amount of work that Juanita does plus the amount of work that Rick does is 9 hours equal to the whole job. This translates to ) to get Now we multiply by the LCD (which is or 6 It doesn t make sense for to be a negative number, so we take 9 as the only legitimate solution. 8. To solve by elimination, we need to make it so that the coefficients of one variable are opposites in the two equations. In this problem, we can accomplish this by multiplying the first equation by on both sides. 8y 6 y6 Then we add the two equations together to get

8 5y 0 y 4 Substituting 4 in for y in either equation will give Thus the solution is,4. 9. This system is already nicely set up to be solved by substitution, since is given in terms of y. We can substitute 8 7y for in the second equation to get y 6y 19 Thus the solution is (7, -5) y 6y y 19 4y 15 y Using the second equation, we can easily solve for y in terms of. y19. Then we substitute this epression in for y in the first equation or Thus the solution set is If 10 then y 9, and if then y ,9,,16. It s a good idea to remember that the first equation is a parabola, and the second equation is a line. So the solution set contains the two points where the line and the parabola intersect. 41. Though we can use substitution to solve this system, it looks like it would be easiest to eliminate y. y z 5 y z 6 y z We can simply add the first two equations together and then add the second and third equations together to get a system of two equations in two variables: z 11 z 9 Now we can subtract the second of these two equations from the first to get a single equation in a single variable: Then we substitute appropriately to find the values of y and z.

9 z 9 6 z 9 z 15 z 5 y 5 5 y 5 y Thus the solution is,, According to what we ve learned in class, the equation y of the circle with center 6, and radius. So the graph looks like 6 4 is the equation y4 4. The equation 1 is the equation of the ellipse with center (, -4), 4 5 horizontal minor ais of length 4, and vertical major ais of length 10. Here s the graph: 44. The function f has a domain of (since these are all the -values that give us a nonnegative radicand). The range is all nonnegative reals. The graph is the same shape as the y graph, just shifted units to the left.

10 f 45. is an eponential function. The domain is all real numbers, and the range is all positive real numbers. We can simply plug in some values for and find the corresponding values for y to get a few ordered pairs and graph the function. For eample, 0, 1, 1, 1, 1, 1,,1,,, and 4,4 4 8 are all points on the graph. To graph the inverse function, we just reverse the coordinates of these points and plot them. Here are the graphs: To find an equation for the inverse, we swap and y in the original equation and solve for y. y log y y log 1 f log 46. Remember that the function f a h k is a V shaped function with verte at hk,, and to the right of the verte it has a slope of a, and to the left of the verte, it has a slope of a. Thus the function f has verte (8, -6) and has a slope of to the right of the verte and a slope of 1 to the left of the verte. The graph is shown at right.

11 y 47. The equation 1 is the 6 9 hyperbola with vertical ais centered at the origin, with vertices at (0,6) and (0,-6). The asymptotes are the lines y. The graph is shown at right. 48. The function f 5 is a greatest integer function. We say that f of is the greatest integer less than or equal to + 5. Thus the graph looks like this: y is the verte form of the equation for the parabola with verte 5 at (5,1). It opens upward and is 5 times as wide as the parabola y. Thus the graph looks like this: 49. The equation Remember that the absolute value of an epression means the distance between that epression and zero. So the given inequality means that is between and. That is,

12 p p8 0. We factor the left side to get p 4 p 0. A quick way to determine the solution set is to realize that the equation y p p 8 is for an upward opening parabola with intercepts at 4 and. So the solution set to the inequality is really the set of all p values for which the parabola is above the horizontal ais. Since the parabola opens upward, these are all the values outside of 4 and. That is,, 4,. We can factor the left side of the inequality to get t t 5. t 4t This makes it easy to see that the zeros of the left side are at 5 and 1. We see that when t is less than 1, the product is positive. When t is between 1 and 5, the product is negative, and when t is greater than 5, the product is positive. Thus the solution set is [-1, 5] Notice that the numerator is always negative, and the denominator is zero when 5 = -/5. The denominator is negative when is greater than /5 but positive when is less 5 than /5. Thus the fraction is positive to the right of /5 but negative to the left of 5 /5 (and undefined at /5). Thus the solution set (i.e., the set of -values for which the inequality is true) is the set of all reals greater than /5.