Chapter 11 and 12 Quiz Review Divide. Identify the divisor, dividend

Chapter 11 and 12 Quiz Review
Divide. Identify the divisor, dividend, quotient, and remainder if any.
1. (12π‘Ÿ 4 βˆ’ 30π‘Ÿ 2 βˆ’ 72π‘Ÿ) ÷ (-6r)
2. (8π‘ž + π‘ž 2 + 7) ÷ (7 + π‘ž)
3. (4𝑀 4 + 3𝑀 βˆ’ 7) ÷ (2𝑀 βˆ’ 1)
Simplify. Find the excluded values.
4.
5.
6.
3𝑐+33
𝑐+11
𝑣+2
𝑣 2 βˆ’4
8π‘₯ 3 βˆ’52π‘₯ 2 +84π‘₯
8π‘₯ 3 βˆ’48π‘₯ 2 +72π‘₯
Solve. Check for extraneous solutions.
7. √3π‘₯ + 4 = 16
8. √9π‘₯ βˆ’ 30 = √4π‘₯ + 5
9. √π‘₯ βˆ’ 8 βˆ’ 4 = βˆ’2
10. √3 βˆ’ π‘₯ = π‘₯ + 1
Find the unknown lengths of the right triangle. Write your answer in simplest radical form.
11.
x√2 𝑖𝑛
x 𝑖𝑛
2x βˆ’ 5 𝑖𝑛
12. A right triangle has sides a=x, b=x+2 and c=14. Find the value of each side.
13. The average late payment fee F (in dollars) on a credit card account during the period 1994-2003 can be
12+1.6π‘₯ 2
modeled by 𝐹 = 1+0.04π‘₯2 where x is the number of years since 1994. Rewrite the model so that it has only
whole number coefficients. Then simplify the model and approximate the average late payment fee in 2003.
Simplify the expression.
14. √15 + 5√3 βˆ’ 2√27
2
15. (3 βˆ’ 2√7)
16.
5
+
√7
2
√14
121
17. √
16π‘š2
4π‘₯ 2
3
18. 6√
1
2
19. (2 βˆ’ 4√3)
80π‘₯ 5
20. √ 15π‘₯
Solutions.
1. βˆ’πŸπ’“πŸ‘ + πŸ“π’“ + 𝟏𝟐 divisor: -6r, dividend: πŸπŸπ’“πŸ’ βˆ’ πŸ‘πŸŽπ’“πŸ βˆ’ πŸ•πŸπ’“, quotient: βˆ’πŸπ’“πŸ‘ + πŸ“π’“ + 𝟏𝟐; no remainder
2. q+1; divisor: q+7; dividend: π’’πŸ + πŸ–π’’ + πŸ•, quotient: q+1; no remainder
𝟏
πŸ•
3. πŸπ’˜πŸ‘ + π’˜πŸ + 𝟐 π’˜ + πŸ’ +
𝟐𝟏
πŸ’
πŸπ’˜βˆ’πŸ
(we will learn to simplify the last term next week); divisor: 2w-1; dividend:
πŸ’π’˜πŸ’ + πŸ‘π’˜ βˆ’ πŸ•; remainder is 21/4
4. 3; c≠11
5.
6.
𝟏
v≠-2, 2
π’—βˆ’πŸ
πŸπ’™βˆ’πŸ•
x≠0,3
πŸπ’™βˆ’πŸ”
7. {48}
8. {12}
9. {7}
10.
βˆ’πŸ‘βˆ“βˆšπŸπŸ•
𝟐
11. Sides are 5 in, 5 in, and πŸ“βˆšπŸ
12. 𝒂 = βˆ’πŸ + βˆšπŸ—πŸ•, 𝒃 = 𝟏 + βˆšπŸ—πŸ•, 𝒄 = πŸπŸ’
13.
πŸ’πŸŽπ’™πŸ +πŸ‘πŸŽπŸŽ
π’™πŸ +πŸπŸ“
14. βˆšπŸπŸ“ βˆ’ βˆšπŸ‘
15. πŸ‘πŸ• βˆ’ πŸπŸβˆšπŸ•
16.
17.
πŸ“βˆšπŸ•+βˆšπŸπŸ’
πŸ“βˆšπŸ•
βˆšπŸπŸ’
or
+
πŸ•
πŸπŸ’
πŸ•
𝟏𝟏
πŸ’π’Ž
18. πŸ’π’™βˆšπŸ‘
𝟏
19. πŸ’πŸ– πŸ’ βˆ’ πŸ’βˆšπŸ‘
20.
πŸ’π’™πŸ βˆšπŸ‘
πŸ‘