Title: M2.2 Formulas and Variation Quiz
Course: Technical Math for Industry > Section: Module 2.2 Formulas and Variation

Q: Find the domain of the rational expression:
- all real numbers except 3 (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 3.)
- all real numbers except 6 (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 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 4 (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 3

Q: Simplify:
A: (a choice with no text)
Feedback: 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.
Feedback: Incorrect. You may have factored the numerator and denominator incorrectly, perhaps ending up with the fraction that simplified to 1. Try factoring again. Separate the like factors from the others, and simplify fractions with equal numerators and denominators to 1.
Feedback: Correct.
Feedback: Incorrect. You may have treated the first terms in the numerator and denominator as if they were factors. You cannot do this. To simplify rational expressions, factor the entire numerator and factor the entire denominator. Separate the like factors from the others, and simplify fractions with equal numerators and denominators to 1.

Q: Multiply. State the product in simplest form.
A: (a choice with no text)
Feedback: Incorrect. Factor the numerators and denominators. Regroup common factors in the same fraction and reduce to 1. Multiply simplified expressions. Be careful not to move factors from the denominator to the numerator (or from the numerator to the denominator).
Feedback: Incorrect. You may have correctly simplified the expression to , but then incorrectly tried to "cancel" the terms. To simplify fractions you have to factor the numerators and denominators. Then regroup common factors in the same fraction and reduce to 1.
Feedback: 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.
Feedback: 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)
Feedback: Incorrect. You may have confused a factor with a factor. Multiply by the reciprocal of the divisor. Factor the numerators and denominators. Regroup common factors in the same fraction and reduce to 1. Multiply simplified expressions.
Feedback: 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.
Feedback: 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.
Feedback: Correct.

Q: Add. State the sum in simplest form.
A: (a choice with no text)
Feedback: 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.
Feedback: Correct. The least common denominator is .
Feedback: Incorrect. It appears that you attempted to rewrite the fractions with the same denominator, but multiplied the numerators by the wrong factors. Start with: Multiply, and then add the numerators.
Feedback: 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.

Q: Subtract. State the difference in simplest form.
A: (a choice with no text)
Feedback: Correct. The least common denominator is . Rewrite each fraction to have this denominator:
Feedback: Incorrect. It appears that you subtracted the numerators and subtracted 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.
Feedback: 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.
Feedback: Incorrect. You may have realized that the the least common denominator is , but then incorrectly rewrote the fractions as: Instead, you should rewrite them using: Multiply, and then subtract the numerators.

Q: Add. State the sum in simplest form.
A: (a choice with no text)
Feedback: Incorrect. It appears that you added the numerators and added the denominators. To add these fractions, you must first find the least common denominator, which in this case is . Then rewrite the first fraction as an equivalent fraction with this denominator. Add the numerators while leaving the denominator the same.
Feedback: Correct. Because , this is the least common denominator.
Feedback: Incorrect. You correctly found the least common denominator. However, you probably made a mistake when rewriting the first fraction, replacing with . Do not forget to use FOIL when multiplying binomials.
Feedback: Incorrect. You found the correct least common denominator, but then just added the numerators. To add these fractions, you must rewrite the first fraction as an equivalent fraction with the denominator . Then add the numerators while leaving the denominator the same.

Q: Perform the indicated operations, then simplify.
A: (a choice with no text)
Feedback: 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.
Feedback: Incorrect. You may have just combined the numerators and then added the denominators. Instead, find the least common denominator and use this to rewrite each term as an equivalent fraction: Then simplify and combine the numerators.
Feedback: Correct.
Feedback: Incorrect. You probably correctly rewrote the fractions using the LCD, but made a mistake simplifying the expression on the last step. When subtracting , be sure to distribute the minus sign to and 90. You added the 90.

Q: Simplify:
A: (a choice with no text)
Feedback: Incorrect. You may have simplified the expression correctly until the very last step, when you placed the 2 from the denominator into the numerator.
Feedback: Correct. Multiply the numerator and denominator by , which is the LCD of all of the terms.
Feedback: 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.
Feedback: 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)
Feedback: Correct. Multiply both sides by the common denominator 14: or , so
Feedback: Incorrect. You may have chosen to multiply both sides by 7 and made a mistake when simplifying the right side. To clear fractions, you should multiply both sides by a common denominator. In this case, 14 will work.
Feedback: Incorrect. You may have correctly chosen to multiply both sides by 14, and canceled correctly, but made a mistake with signs when solving the resulting equation.
Feedback: Incorrect. You may have correctly chosen to multiply both sides by 14, but later made a mistake by failing to distribute on the left side.

Q: Solve the equation:

What are the solutions?
- only (Correct. Note that and 1 are excluded values. Multiply both sides by the common denominator : Simplify: Distribute: Solve for x: , ,)
A: only
Feedback: 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.
Feedback: 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.
Feedback: 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 factors 6 and 4.

Q: The Westminster Widget Company has an old machine that can produce 50 widgets in five hours. Now they have purchased a new machine that can produce 30 widgets in two hours. Working together, how long will it take the two machines to produce 100 widgets?
- hour (Incorrect. You probably set up a rate incorrectly. The old machine's rate is or 10 widgets per hour. The new machine's rate is or 15 widgets per hour. Find their combined rate. Find a fraction that represents the rate for the new job in widgets per hour. Set these two numbers equal and solve the equation.)
- 4 hours (Correct. The old machine's rate is or 10 widgets per hour. The new machine's rate is or 15 widgets per hour. Their combined rate is 25 widgets per hour. So, if the amount of time, in hours, needed for the two machines to produce 100 widgets, then: , so , and)
- hours (Incorrect. You may have tried to find the combined rate of the machines by adding the number of widgets and adding the times and then dividing. The old machine's rate is or 10 widgets per hour. The new machine's rate is or 15 widgets per hour. You need to add these to find their combined rate.)
- hours (Incorrect. You may have incorrectly tried to find rates by dividing time by work, instead of the other way around. The old machine's rate is or 10 widgets per hour. The new machine's rate is or 15 widgets per hour. Their combined rate equals 100 widgets divided by the time it takes.)
A: 4 hours

Q: The formula for the volume of a pyramid is , where B is the area of the base and h is the height. Rearrange the formula to solve for the height (h).
A: (a choice with no text)
Feedback: Correct. Multiply both sides by 3: Simplify: Divide both sides by B: Simplify:
Feedback: Incorrect. To rewrite any equation, do one operation at a time and make sure that you do the same thing to both sides. To start this problem, you can multiply both sides by 3.
Feedback: Incorrect. To rewrite any equation, do one operation at a time and make sure that you do the same thing to both sides. To start this problem, you can multiply both sides by 3.
Feedback: Incorrect. To rewrite any equation, do one operation at a time and make sure that you do the same thing to both sides. To start this problem, you can multiply both sides by 3.

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

I

II

III
- III only (Correct. In a joint variation, one variable depends on two or more other variables. Because the value of A depends on two variables in formula III, and only one variable in the other formulas, only III shows a joint variation.)
- 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 A depends will tell you if a formula shows a joint variation.)
- II and III (Incorrect. Formula III does show a joint variation because A depends on two variables. However, formula II does not. You may have forgotten that is a constant and counted it as a variable. In formula II, A depends on only one variable.)
- II only (Incorrect. You may have forgotten that is a constant and counted it as a variable. In formula II, A depends on only one variable, so it is not a joint variation.)
A: III only

Q: The value of y varies jointly with the values of x and z. When and , the value of y is  360.

What is the value of y when and ?
- 600 (Correct. Use the equation for joint variation and the given values for x, y, and z to find k. , so Use this value of k and the given second set of values for x and z to find y: or)
- 10 (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 joint variation equation , along with the second values for x and z. Solve for y.)
- 450 (Incorrect. You may have used a direct variation equation instead of the joint variation equation, ignoring the value. Be sure to use the joint variation equation .)
- 170 (Incorrect. You may have correctly found that the constant , and incorrectly solved the second equation, , possibly by adding the 5 and 12.)
A: 600

Q: The graph represents distance traveled varying directly with time.

What would be the distance traveled after 9 hours?
- 40 miles (Incorrect. You probably found the constant of variation and stopped there. The graph shows that 200 miles are traveled in 5 hours. Because this is a direct variation problem, solve for k, the constant of variation, using the equation , where and . Then use the k value in a new direct variation equation, with , and solve for y.)
- 400 miles (Incorrect. You may have misread the graph. The graph shows that 200 miles are traveled in 5 hours. Because this is a direct variation problem, solve for k, the constant of variation, using the equation , where and . Then use the k value in a new direct variation equation, with , and solve for y.)
- 360 miles (Correct. The graph shows that 200 miles are traveled in 5 hours. Using the equation for direct variation, , you get: . Solve for k to find . Use this k value and in a new equation for direct variation, and solve for y: or)
- 450 miles (Incorrect. You may have misread the graph. The graph shows that 200 miles are traveled in 5 hours. Because this is a direct variation problem, solve for k, the constant of variation, using the equation , where and . Then use the k value in a new direct variation equation, with , and solve for y.)
A: 360 miles

Q: Shirley is drawing triangles that have the same area. The base of each triangle varies inversely with the height. What are the possible base and height of a second triangle if the first triangle's base is 12 and its height is 8?
- 60 and 36 (Incorrect. You may have found the correct value of k, but then chose the two measurements whose sum is 96. However, the numbers need to satisfy , an inverse variation equation. For the first triangle, , so . Substitute 60 and 36. does not yield , so this is not an inverse relationship.)
- 120 and 80 (Incorrect. This would be the correct answer for a direct proportion. However, the numbers need to satisfy , an inverse variation equation. For the first triangle, , so . Substitute 120 and 80. does not yield , so this is not an inverse relationship.)
- 16 and 6 (Correct. Use the inverse variation equation : , so . Substitute 16 and 6: also yields , so this is an inverse relationship.)
- 10 and 10 (Incorrect. You may have chosen this answer because the sum of the two numbers is the same as the sum of the two numbers in the measurements of the first triangle. However, the numbers need to satisfy , an inverse variation equation. For the first triangle, , so . Substitute 10 and 10. does not yield , so this is not an inverse relationship.)
A: 16 and 6

Q: The volume of a cylinder varies jointly with the square of its radius and with its height:

Cylinder A has a volume of 254.34 cubic inches and has a radius of 3 inches and a height of 9 inches. What is the volume of cylinder B, which has a radius of 4 inches and a height of 5 inches?
- 251.2 cubic inches (Correct. Using the joint variation equation , substitute the values for cylinder A. Use 3.14 in the next joint variation equation to find the volume of cylinder B.)
- 3.14 cubic inches (Incorrect. You may have found only k, the constant of variation. Be sure to complete the problem by substituting , , and in the equation .)
- 62.8 cubic inches (Incorrect. You may have found the correct value of k, by using and solving , but you incorrectly solved the second equation by not squaring the radius r, which is 4.)
- 188.4 cubic inches (Incorrect. You may have done your calculations without squaring the radius. When you substitute into , you should first get . Then you find k. When you are finding the volume of B, make sure that you square the radius r, which is 4.)
A: 251.2 cubic inches