- Find the indicated root \sqrt[3]{-64}
- -4
- -16
- 16
- 4
- Cannot be evaluated
- 3^{3x}=9^{x-1}
- 2
- -2
- \frac{1}{2}
- \frac{1}{3}
- 3
- 3\sqrt{20}+6\sqrt{45}
- 24\sqrt{5}
- 9\sqrt{65}
- 3\sqrt{5}
- 20\sqrt{5}
- 18\sqrt{5}
- 2a^{2}-3(b-c)^{2}
- [\sqrt{2}a-\sqrt{3}b+\sqrt{3}c][\sqrt{2}a+\sqrt{3}b-\sqrt{3}c]
- [\sqrt{2}a-\sqrt{3}b+\sqrt{3}c][\sqrt{2}a-\sqrt{3}b+\sqrt{3}c]
- [\sqrt{2}a-\sqrt{3}b.\sqrt{3}c][\sqrt{2}a+\sqrt{3}b.\sqrt{3}c]
- [2a-3b+3c][2a+3b-3c]
- [\sqrt{2}a+\sqrt{3}b+\sqrt{3}c][\sqrt{2}a+\sqrt{3}b+\sqrt{3}c]
- (2\sqrt{5}+5\sqrt{2})(5\sqrt{5}-2\sqrt{2})
- 30-21\sqrt{10}
- 30+21\sqrt{10}
- -30+21\sqrt{10}
- -10
- +10
- x^{3}-36x=0
- -6 and 6
- -3, 0 and 3
- -6, 0 and 6
- 36
- 6
- g(x)=-2x^{2}+5x-1
g(-5)
- log_{25}(\frac{1}{125})=
- \frac{-3}{2}
- \frac{2}{3}
- \frac{3}{2}
- -3
- 2
- If log_{b}(5)=0.81 and log_{b}(3)=0.13 find log_{b}(15)
- 0.1053
- 0.94
- 0.10
- 0.9
- 0.95
- If log_{5}N=2, find N
- \frac{5}{2}
- 10
- \frac{2}{5}
- log_{2}5
- 25
This site prepare students for math exams, SAT, GRE, GMAT,CLEP practice test, CLEP college Algebra,CLEP precalculus,CLEP Mathematics
Tuesday, June 28, 2011
CLEP College Algebra Practice Questions - 5
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment