Title: Unit 15 Quiz
Course: Technical Math for Industry

Q: Find the domain of the rational expression:
- all real numbers except 0 (Incorrect. You may not have realized that, though the denominator cannot equal 0, this does not necessarily mean that x cannot equal 0. Set equal to 0 and solve the equation. This will give you the excluded value.)
- all real numbers except (Incorrect. You may have realized that the denominator could not be equal to 0, but you made a mistake finding the excluded value. You need to solve the equation to find the excluded value.)
- all real numbers except (Correct. For rational expressions, the domain will exclude values of x that make the denominator 0. Set the denominator equal to 0 and solve for x. , so . The domain is all real numbers except .)
- all real numbers except 8 (Incorrect. You may have set the numerator equal to 0 and then solved. Recall that dividing by 0 is undefined. Therefore, to find the excluded values for a rational expression, you need to set the denominator equal to 0 and then solve.)
A: all real numbers except

Q: Simplify:
A: (a choice with no text)
If wrong: Incorrect. You may have correctly factored, separated like factors, and simplified a fraction with equal numerator and denominator to 1. However, you probably tried to simplify further when you did not have common factors, but had common terms of x in the numerator and denominator. These cannot be removed.
If wrong: Incorrect. You may have incorrectly factored both the numerator and denominator, obtaining a common factor of . Be sure to check your factoring by multiplying out the factorization to be sure you obtain the given polynomial.
If right: Correct.
If wrong: Incorrect. You may have ignored the extra x in the denominator when you factored the numerator and denominator.

Q: Multiply. State the product in simplest form.
A: (a choice with no text)
If wrong: Incorrect. You may have tried to factor out term for term, factoring each a variable in the numerator with each a variable in the denominator. To simplify fractions you need to factor the numerators and denominators. Then regroup common factors in the same fraction and reduce to 1.
If wrong: Incorrect. This answer is equivalent to the given expression, but it can be further simplified because the numerator and denominator have a common factor. Factor the denominator, regroup common factors in the same fraction, and reduce to 1.
If wrong: Incorrect. You may have dropped or forgotten about a couple of the factors. Factor the numerators and denominators. Regroup common factors in the same fraction and reduce to 1. Multiply simplified expressions. Be careful to include all variables or constants that do not have a common factor and reduce to 1.
If right: Correct. Factor the numerators and denominators. Regroup common factors in the same fraction and reduce to 1. Multiply simplified expressions.

Q: Divide and state the quotient in simplest form.
A: (a choice with no text)
If wrong: Incorrect. You probably correctly multiplied by the reciprocal of the divisor and then started to simplify. However, this answer is not in simplest form. You need to factor both quadratic expressions, regroup common factors in the same fraction, and reduce to 1.
If right: Correct.
If wrong: Incorrect. You need to multiply by the reciprocal of the divisor. It appears that you used the reciprocal of the first fraction instead. You probably did the correct factoring and simplifying after that.
If wrong: Incorrect. You may have correctly multiplied by the reciprocal of the divisor and started to simplify correctly, but incorrectly "canceled" like x terms. It is only common factors in the numerator and denominator, not terms, that will "cancel" when you put them in the same fraction and reduce.

Q: Add. State the sum in simplest form.
A: (a choice with no text)
If right: Correct. The least common denominator is .
If wrong: Incorrect. You may have added the numerators and added the denominators. To add these fractions, rewrite each with the least common denominator, which is . Start with: Simplify the numerators and add them.
If wrong: Incorrect. It appears that you attempted to rewrite the fractions with the same denominator, but multiplied the numerators by the wrong factors. Start with: Simplify the numerators and add them.
If wrong: Incorrect. You may have added the numerators and multiplied the denominators. To add these fractions, rewrite each with the least common denominator, which is . Start with: Simplify the numerators and add them.

Q: Subtract. State the difference in simplest form.
A: (a choice with no text)
If wrong: Incorrect. It appears that you rewrote the fractions correctly, but then you subtracted incorrectly. When you simplify the expression , remember to distribute the minus sign on the outside to the term .
If wrong: Incorrect. It appears that you subtracted the numerators and multiplied the denominators. Instead, you need to find the least common denominator, which is . Then rewrite both fractions so that they have this denominator. Finally, subtract only the numerators while keeping the denominator the same.
If right: Correct. The least common denominator is . So
If wrong: Incorrect. You may have correctly multiplied the first fraction by and the second by , but then simplified the numerators incorrectly. You need to use the distributive property when simplifying and .

Q: Add. State the sum in simplest form.
A: (a choice with no text)
If wrong: Incorrect. You found a common denominator by multiplying the two denominators, but then you just added the numerators. To add fractions you first need to find the least common multiple of the denominators. For these fractions it is . Then rewrite both fractions with this denominator. Add the numerators.
If right: Correct. and , so the LCM
If wrong: Incorrect. It appears that you added the numerators and the denominators. To add fractions you must first find the least common multiple of the denominators. For these fractions it is . Then rewrite both fractions with this denominator. Add the numerators.
If wrong: Incorrect. The sum is correct, but it is not in simplest form. You can factor out a common factor of n in the numerator and denominator, then reduce to 1.

Q: Perform the indicated operations, then simplify.
A: (a choice with no text)
If wrong: Incorrect. You may have correctly rewritten the terms as equivalent fractions with least common denominator and combined the numerators while keeping the denominator the same. However, you cannot cancel a term of 36 on the top with a on the bottom. You need common factors to simplify.
If right: Correct.
If wrong: Incorrect. Though you may have correctly rewritten the terms as equivalent fractions with least common denominator , when attempting to combine the numerators you probably forgot to distribute the minus sign in the second term: After all numerators are simplified, you may combine them while keeping the denominator the same.
If wrong: Incorrect. It appears that you combined the original numerators while changing the denominators to the correct LCD. Instead, rewrite each term as an equivalent fraction before adding or subtracting: Then simplify and combine the numerators.

Q: Simplify:
A: (a choice with no text)
If right: Correct. Multiply the numerator and denominator by , which is the LCD of all of the terms.
If wrong: Incorrect. You may have tried to divide term by term, but this is the incorrect order of operations. Multiply the numerator and denominator by the LCD of all of the terms: Distribute the and simplify. Then factor the numerator and the denominator and identify any common factors to finish simplifying.
If wrong: Incorrect. You may have simplified the expression correctly until the very last step, when you multiplied the 5 from the denominator with the numerator.
If wrong: Incorrect. You correctly simplified to an ordinary fraction, but it is not in simplest form. Be sure to factor the numerator and the denominator and identify any common factors.

Q: Solve the equation:
A: (a choice with no text)
If wrong: Incorrect. You may have correctly chosen to multiply both sides by 16, and canceled correctly, but made a mistake with signs when solving the resulting equation.
If wrong: Incorrect. You may have correctly chosen to multiply both sides by 16, but later made a mistake by failing to distribute on the left side.
If right: Correct. Multiply both sides by the common denominator 16: or , so
If wrong: Incorrect. You may have chosen to multiply both sides by 8 and made a mistake when simplifying the right side. To clear fractions, you should multiply both sides by a common denominator. In this case, 16 will work.

Q: Solve the equation:

What are the solutions?
- only (Correct. Note that and 6 are excluded values. Multiply both sides by the common denominator : Simplify: Distribute: Solve for x: , ,)
A: only
If wrong: Incorrect. You may have tried to multiply both sides by the common denominator to clear the fractions, but ended up with the incorrect equation . When you have a fraction such as , it is the common factor in the numerator and denominator that will cancel, because simplifies to 1.
If wrong: Incorrect. These are the excluded values, not the solutions to the equation. Multiply both sides by the common denominator to clear the denominators. Use the distributive property and solve for x.
If wrong: Incorrect. You may have correctly multiplied both sides by the common denominator and cleared the fractions. But then you probably forgot to distribute the constant factor 12.

Q: A shoe repairman is working with his assistant, who takes 1.5 times as long to repair a pair of shoes. Together they can fix 20 pairs of shoes in an eight-hour day. How long does it take the repairman to fix one pair of shoes by himself?
- 1 hour (Incorrect. You may have set up and solved the equation correctly, but this is the amount of time that it takes the assistant to fix a pair of shoes.)
- hour (Correct. Let the time, in hours, it takes the repairman to fix one pair of shoes. Then the amount of time it takes the assistant. Their rates are and pairs per hour. Their combined rate is pairs per hour. So or Therefore, hour, so hour.)
- hour (Incorrect. Let the time, in hours, it takes the repairman to fix one pair of shoes. Then the amount of time it takes the assistant. Their rates are and pairs per hour. You may have added these fractions incorrectly when finding their combined rate. To add two fractions, you must rewrite one or both so that they have a common denominator.)
- hour (Incorrect. You may have incorrectly found your rates by using time divided by work. Let the time, in hours, it takes the repairman to fix one pair of shoes. Then the amount of time it takes the assistant. Their rates are and pairs per hour. Add these rates to get their combined rate. Then use rate times time equals amount of work done.)
A: hour

Q: Solve for k.
A: (a choice with no text)
If wrong: Incorrect. When isolating the variable, make sure that you perform the same operation on both sides of the equation. For example, if you subtract from the right side, you must also subtract from the left side. If you divide both sides by 3, you must divide every term by 3.
If wrong: Incorrect. When isolating the variable, make sure that you perform the same operation on both sides of the equation. For example, if you subtract from the right side, you must also subtract from the left side.
If wrong: Incorrect. When isolating the variable, make sure that you perform the same operation on both sides of the equation. For example, you can divide both sides by 3. If you do this, you must divide every term by 3.
If right: Correct. Subtract from both sides: Divide both sides by 3: Simplify:

Q: Which volume formula or formulas show(s) a joint variation?

I

II

III
- I, II, and III (Incorrect. In a joint variation, one variable depends on two or more other variables. Counting the number of different variables on which V depends will tell you if a formula shows a joint variation.)
- III only (Incorrect. Formula III does show a joint variation because V depends on two variables, but it is not the only one. Remember to look for formulas in which V depends on two or more variables.)
- I only (Incorrect. You may have thought that because the variable was cubed it was a joint variation. In formula I, V depends on only one variable, so it is not a joint variation.)
- II and III only (Correct. In a joint variation, one variable depends on two or more other variables. Because the value of V depends on two variables in formulas II and III, and only one variable in the other formula, both II and III show a joint variation.)
A: II and III only

Q: The value of y is inversely proportional to the value of x. When , .

What is the value of y when ?
- 28 (Correct. Use the equation for inverse variation and the given values for x and y to find k. so Use this value of k and the given second value for x to find y: or)
- 45 (Incorrect. You may have misunderstood inverse variation and subtracted the x and y values, obtaining 36 to add to 9. Use the equation for inverse variation and the given values for x and y to find k. Then use this value of k and the given second value for x to find y.)
- 63 (Incorrect. You may have used the direct variation equation. Use the equation for inverse variation and the given values for x and y to find k. Then use this value of k and the given second value for x to find y.)
- 252 (Incorrect. You may have found only the value of the constant k, in which case you did not complete the problem. Substitute the value for k in a second inverse variation equation , along with the second value for x. Solve for y.)
A: 28

Q: The total cost of an item including sales tax is directly proportional to its price. If the total cost of a $24 item is $25.44, what is the total cost on a $75 item?
- $76.44 (Incorrect. You may have found the amount of tax on the $24 item ($1.44) and then added that to $75. Because this is a direct variation problem, use the equation , where y is the total cost and x is the item's price. Substitute and : . Solve for k. Then use this value of k and in a new direct variation equation, and solve for y.)
- $79.50 (Correct. Because this is a direct variation problem, use the equation , where and , and solve for k. , so Use the k value in a new direct variation equation in which , and solve for y:)
- $106 (Incorrect. It appears that you found the value for k, and then you may have changed it to a percent to obtain the 106. You need to use the value of k, which is 1.06, and in a new direct variation equation, and solve for y.)
- $795 (Incorrect. You probably made a mistake with the placement of a decimal. Because this is a direct variation problem, use the equation , where y is the total cost and x is the item's price. Substitute and : . Solve for k. Then use this value of k and in a new direct variation equation, and solve for y.)
A: $79.50

Q: Mr. Jones jogs the same route each day. The amount of time he jogs is inversely proportional to his jogging rate.

What option gives possible rates and times for two of his jogs?
- 4 mph for 2.25 hours and 6 mph for 1.5 hours (Correct. Use an inverse variation equation: or . The first two numbers give you , so k would be 9. The second two numbers give you . Because k is the same, the relationship is inversely proportional.)
- 5 mph for 2 hours and 4 mph for 3 hours (Incorrect. You may have thought that the sums of the two numbers should be constant. However, these numbers need to satisfy an inverse variation equation: or . The first two numbers give you , while the second two numbers give you . Because k is not the same, the relationship is not inversely proportional.)
- 4.5 mph for 3 hours and 6 mph for 4 hours (Incorrect. These numbers satisfy a directly proportional relationship. They need to satisfy an inverse variation equation: or . The first two numbers give you , while the second two numbers give you . Because k is not the same, the relationship is not inversely proportional.)
- 6 mph for 1.5 hours and 5 mph for 1.25 hours (Incorrect. These numbers satisfy a directly proportional relationship. They need to satisfy an inverse variation equation: or . The first two numbers give you , while the second two numbers give you . Because k is not the same, the relationship is not inversely proportional.)
A: 4 mph for 2.25 hours and 6 mph for 1.5 hours

Q: The area of a triangle varies jointly with the height of the triangle and the length of its base. The area of one triangle is 200 square centimeters when its height is 25 centimeters and its base length is 16 centimeters. What is the area of a triangle having a height of 20 centimeters and a base length of 15 centimeters?
- 160 square centimeters (Incorrect. You may have divided 200 by 25 and used that 8 to multiply by 20. Be sure to use the joint variation equation to find the constant of variation, k. Then use that value of k along with the second set of values for x and z in a second joint variation equation to solve for y.)
- 300 square centimeters (Incorrect. You may just have multiplied 20 by 15. Be sure to use the joint variation equation to find the constant of variation, k. Then use that value of k, along with the second set of values for x and z, in a second joint variation equation to solve for y.)
- 600 square centimeters (Incorrect. You may have found the correct value of k, but then instead of multiplying the product of 20 and 15 by , you may have multiplied the product by 2. The correct calculation would be .)
- 150 square centimeters (Correct. Use a joint variation equation, : , so Substitute k in a new joint variation equation using and : , so)
A: 150 square centimeters