Part B (solution): For 0.42... (with the two repeating forever), we need to break up the decimal into the part that does not repeat and the part that repeats.
For the part that does not repeat, pretend that the repeating part is not there (so you have 0.4). Then think about place value. The non-repeating part of this decimal (0.4) is 4 tenths, so write that as a fraction: 4/10.
For the part that repeats, pretend the part that does not repeat is not there (so you have 0.2). Use the rule of 9s, which tells you to put the repeating part (2) over a nine for every place value that is repeating. In this case there is only one repeating place value (again, 2), so we have 2/9. However, since we skipped a place value (the 4 in our given problem), we need to multiply that place value (in this case 1/10) by 2/9, which is 2/90. Finally, we need to add that to 4/10. Make sure your answer is simplified. Click HERE for help on Question 1.
Part C (solution): For 0.42... (with the four AND two repeating forever), use the rule of 9s. The rule of 9s tells you to put the repeating part (42) over a nine for every place value that is repeating. In this case there is two repeating place values, so we have 42/99. Make sure your answer is simplified. Click HERE for help on Question 1.
Parts A-C (work shown): Please make sure you include all of your work. Click HERE for help on Question 1.
Question 2
Part A (explanation): Please explain what this student did wrong. What was his error? It might be helpful to estimate √75 the way you know how. That should help you see what this student did wrong. Click HERE for help on Question 2.
Part B (solution): Estimate the value of √75 to the nearest tenth. 75 is between the perfect squares 64 and 81. So √75 is between 8 and 9. The distance from 64 to 75 is 11. The distance from 64 to 81 is 17. The ratio is 11/17, or about .647. So, √75 is about .647 bigger than 8. Click HERE for help on Question 2.
Question 3
Part A - (work for π): Approximate the value of π to the nearest tenth. Click HERE for help on Question 3.
Part A - (work for √3): Approximate the value of √3 to the nearest tenth. 3 is between the perfect squares 1 and 4. So √3 is between 1 and 2. The distance from 1 to 3 is 2. The distance from 1 to 4 is 3. The ratio is 2/3, or about .7. So, √3 is about .7 bigger than 1. Click HERE for help on Question 3.
Part A - (work for √5): Approximate the value of √5 to the nearest tenth. 5 is between the perfect squares 4 and 9. So √5 is between 2 and 3. The distance from 4 to 5 is 1. The distance from 4 to 9 is 5. The ratio is 1/5, or .2. So, √5 is about .2 bigger than 2. Click HERE for help on Question 3.
Part A - (work for 2√5): Approximate the value of 2√5 to the nearest tenth. 5 is between the perfect squares 4 and 9. So √5 is between 2 and 3. The distance from 4 to 5 is 1. The distance from 4 to 9 is 5. The ratio is 1/5, or .2. So, √5 is about .2 bigger than 2. Now multiply that value by 2. Click HERE for help on Question 3.
