Proportions And Exponents Quiz

1. Joseph drives 135 miles in 2.5 hours. At the same rate, how far will he be able to travel in 6 hours?

810

324

125

300

Joseph drives at a constant rate, so we can use the formula distance = rate × time. We are given that Joseph drives 135 miles in 2.5 hours, so we can calculate his rate as 135 miles / 2.5 hours = 54 miles per hour. Now we can use this rate to find how far Joseph will be able to travel in 6 hours. distance = rate × time = 54 miles/hour × 6 hours = 324 miles. Therefore, Joseph will be able to travel 324 miles in 6 hours.

Explanation

Joseph drives at a constant rate, so we can use the formula distance = rate × time. We are given that Joseph drives 135 miles in 2.5 hours, so we can calculate his rate as 135 miles / 2.5 hours = 54 miles per hour. Now we can use this rate to find how far Joseph will be able to travel in 6 hours. distance = rate × time = 54 miles/hour × 6 hours = 324 miles. Therefore, Joseph will be able to travel 324 miles in 6 hours.

2. Solve

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mo»(«/mo»«mn»7«/mn»«msup»«mi»x«/mi»«mn»9«/mn»«/msup»«msup»«mi»y«/mi»«mn»4«/mn»«/msup»«msup»«mo»)«/mo»«mn»2«/mn»«/msup»«/math»

not-available-via-ai

Explanation

not-available-via-ai

3. An employee working at an electronics store earned $3582 for working 3 months during the summer. How much could the employee make if they worked for 8 months?

1343.25

149.25

85968

9552

If the employee earned $3582 for working 3 months, we can calculate their monthly earnings by dividing the total earnings by the number of months worked. $3582 divided by 3 is equal to $1194. Therefore, if the employee worked for 8 months, they could make $1194 per month. Multiplying $1194 by 8 gives us $9552, which is the amount the employee could make if they worked for 8 months.

Explanation

If the employee earned $3582 for working 3 months, we can calculate their monthly earnings by dividing the total earnings by the number of months worked. $3582 divided by 3 is equal to $1194. Therefore, if the employee worked for 8 months, they could make $1194 per month. Multiplying $1194 by 8 gives us $9552, which is the amount the employee could make if they worked for 8 months.

4. If 3 pounds of apples costs $0.90, how much will 10 pounds cost?

3

27

2.70

3.30

The cost of apples is directly proportional to the weight. Therefore, if 3 pounds of apples cost $0.90, we can calculate the cost of 10 pounds by finding the ratio of weights. The ratio of 10 pounds to 3 pounds is 10/3. Multiplying this ratio by the cost of 3 pounds, which is $0.90, we get the cost of 10 pounds of apples, which is $3.

Explanation

The cost of apples is directly proportional to the weight. Therefore, if 3 pounds of apples cost $0.90, we can calculate the cost of 10 pounds by finding the ratio of weights. The ratio of 10 pounds to 3 pounds is 10/3. Multiplying this ratio by the cost of 3 pounds, which is $0.90, we get the cost of 10 pounds of apples, which is $3.

5. Solve

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msup»«mn»4«/mn»«mn»2«/mn»«/msup»«mo»*«/mo»«msup»«mn»4«/mn»«mn»10«/mn»«/msup»«/math»

not-available-via-ai

Explanation

not-available-via-ai

6. Solve

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«msup»«mi»x«/mi»«mn»3«/mn»«/msup»«msup»«mi»y«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«mo»§nbsp;«/mo»«mo»*«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«msup»«mi»x«/mi»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/msup»«msup»«mi»y«/mi»«mn»3«/mn»«/msup»«/math»

12x

7xy

The given expression is the product of 12x and 7xy. To simplify the expression, we can multiply the coefficients (12 and 7) to get 84. Then, we can multiply the variables (x and xy) by adding their exponents. The exponent of x is 1, and the exponent of xy is 1+1=2. So, the simplified expression is 84x^2y.

Explanation

The given expression is the product of 12x and 7xy. To simplify the expression, we can multiply the coefficients (12 and 7) to get 84. Then, we can multiply the variables (x and xy) by adding their exponents. The exponent of x is 1, and the exponent of xy is 1+1=2. So, the simplified expression is 84x^2y.

7. In a history class, there are 6 boys for every 5 girls. How many girls are there if there are 3600 boys?

120

3000

4320

6600

Given that there are 6 boys for every 5 girls, we can set up a proportion to find the number of girls. If there are 3600 boys, we can write the proportion as 6/5 = 3600/x, where x represents the number of girls. Cross-multiplying, we get 6x = 5 * 3600, which simplifies to 6x = 18000. Dividing both sides by 6, we find that x = 3000. Therefore, there are 3000 girls in the history class.

Explanation

Given that there are 6 boys for every 5 girls, we can set up a proportion to find the number of girls. If there are 3600 boys, we can write the proportion as 6/5 = 3600/x, where x represents the number of girls. Cross-multiplying, we get 6x = 5 * 3600, which simplifies to 6x = 18000. Dividing both sides by 6, we find that x = 3000. Therefore, there are 3000 girls in the history class.

8. Solve

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mo»(«/mo»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«msup»«mi»y«/mi»«mrow»«mo»-«/mo»«mn»4«/mn»«/mrow»«/msup»«msup»«mo»)«/mo»«mn»0«/mn»«/msup»«/math»

Xy

1

None of the above

The given expression is "xy". When we multiply any number by 1, the value remains the same. Therefore, when we multiply "xy" by 1, it remains as "xy". Hence, the correct answer is 1.

Explanation

The given expression is "xy". When we multiply any number by 1, the value remains the same. Therefore, when we multiply "xy" by 1, it remains as "xy". Hence, the correct answer is 1.

9. Solve

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«msup»«mi»m«/mi»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«mo»*«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math»

5m

not-available-via-ai

Explanation

not-available-via-ai

10. Solve $«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»12«/mn»«msup»«mi»m«/mi»«mn»4«/mn»«/msup»«msup»«mi»n«/mi»«mn»3«/mn»«/msup»«msup»«mi»p«/mi»«mn»3«/mn»«/msup»«/mrow»«mrow»«mn»3«/mn»«msup»«mi»m«/mi»«mn»2«/mn»«/msup»«msup»«mi»n«/mi»«mn»2«/mn»«/msup»«msup»«mi»p«/mi»«mn»4«/mn»«/msup»«/mrow»«/mfrac»«/math»$

not-available-via-ai

Explanation

not-available-via-ai

11. Solve
*Hint: Make all exponents positive.

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mo»-«/mo»«mn»6«/mn»«msup»«mi»x«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«msup»«mi»y«/mi»«mn»0«/mn»«/msup»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/math»

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mo»-«/mo»«mn»6«/mn»«msup»«mi»x«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«msup»«mi»y«/mi»«mn»0«/mn»«/msup»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/math»

not-available-via-ai

Explanation

not-available-via-ai

12. Solve $«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mfrac»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«msup»«mi»x«/mi»«mn»10«/mn»«/msup»«/mfrac»«/math»$