COLLEGE ALGEBRA II (MATH 010)
FALL 2010
FINAL EXAMINATION(Solutions)
1. If log3 2 = A, log3 5 = B, and log3 7 = C write the following
in terms of A, B and C.
b) log3 140
a) log3 35
3
Solution:
(a) log3 35
3 = log3 35 − log3 3 = log3 (5 × 7) − 1
= log3 5 + log3 7 − 1 = B + C − 1
(b) log3 140 = log3 (22 × 5 × 7) = 2 log3 2 + log3 5 + log3 7
= 2A + B + C
2. Solveh fori x:
logx 625
16 = 4
Solution:
x4 = ( 52 )4
x = 52
A
3. Let logb A = 2.227 and logb B = 1.365. Find logb B
.
Solution:
A
logb B
= logb A − logb B = 2.227 − 1.365 = .862
4. Solve for x: log5 (x2 + 1) = 2
Solution:
x2 + 1 = 52
x2 = 24
√
√
x = ± 24 = ±2 6
1
5. Suppose Javonia has $1000 to invest at 8% per year compounded quarterly, how long will it take before she has
$1150?
Solution:
4t
1150 = 1000(1 + .08
4 )
1150 = 1000(1.02)4t
1150
4t
1000 = (1.02)
23
4t
20 = (1.02) h
i
4t
ln( 23
)
=
ln
(1.02)
20
ln( 23
20 ) = 4t ln(1.02)
t=
ln( 23
20 )
4 ln(1.02)
= 1.7644 years
6. Since 1950, the growth in world population in millions closely
fits the function A(t) = 2600e0.018t , where t is the number
of years since 1950. Estimate the population in the year
2003.
Solution:
t = 2003 − 1950 = 53
A(53) = 2600e0.018(53) ≈ 6750 million
2
7. (a) Graph the region that satisfies the constraints:
x ≥ 0, y ≥ 0, x ≤ 3, y ≤ 4, 5x + 4y ≥ 20
(b) Find the maximum and minimum values of the objective function
z = 10x − 8y + 28
subject to the constraints in part (a) and indicate the
coordinates x and y at which they occur.
Solution:
6
5
( 45 , 4)
4
(3, 4)
3
2
(3, 45 )
1
0
0
1
2
3
4
5
6
Figure 1: Region satisfying the constraints
(a)
Point
( 4 , 4)
(b) 5 5
(3, 4 )
(3, 4)
z = 10x − 8y + 28 Max/Min
4
Min
48
Max
26
3
8. For the arithmetic sequence an , find the common difference
and write out the first six terms.
an = 23 + n4
Solution:
2
n
d = an+1 − an = ( 23 + n+1
4 ) − (3 + 4 )
n
1
= n+1
4 − 4 = 4
1
14
a1 = 23 + 41 = 11
a2 = 11
12
12 + 4 = 12
1
17
17
1
20
a3 = 14
12 + 4 = 12 a4 = 12 + 4 = 12
1
23
23
1
26
a5 = 20
12 + 4 = 12 a6 = 12 + 4 = 12
9. Evaluate 5k=1 k 3 − 9.
P
Note: This is different from 5k=1 (k 3 − 9)
Solution:
(13 + 23 + 33 + 43 + 53 ) − 9 = (1 + 8 + 27 + 64 + 125) − 9 = 216
P
10. For each sequence below, determine if the sequence is arithmetic, geometric, or neither. If arithmetic, find the common difference and 67-th term. If geometric, find the common ratio and 8-th term.
3 −3
(a) 2, −2, −6, −10, ...
(b) −3
2 , 4 , 8 , ...
Solution:
(a) d = −4,
an = 2 − 4(n − 1),
a67 = 2 − 4(66) = −262
(b) r = − 21 ,
an = (− 23 )(− 12 )n−1 ,
a8 = (− 23 )(− 12 )7 =
11. Find the sum of the sequence whose general term is
n−1
an = 3 4 n
Solution:
This is an infinite sum with a1 = 14 and r = 43 . Hence
S=
1
4
1− 34
=1
4
3
256
12. In how many ways can a committee consisting of 3 faculty
members and 5 students be formed if 7 faculty members
and 12 students are eligible to serve on the committee?
Solution:
7!
12!
C(7, 3) × C(12, 5) = 3!4!
× 5!7!
= 27, 720
13. On the
14. For each equation shown, indicate whether it is the equation
of a circle, parabola, ellipse, hyperbola, or non-conic.
(a) 8x2 + 3y 2 − 7x + 13 = 0
(b) 6x2 − 6y 2 + 5x − 2y = 0
(c) 3x2 + 8x + 9y + 21 = 0
(d) 9x2 + 9y 2 − 4x + y − 7 = 0
Solution:(Use Ax2 + Cy 2 + Dx + Ey + F = 0)
(a) A = 8, C = 3; AC = 24 > 0 Ellipse
(b) A = 6, C = −6; AC = −36 < 0 Hyperbola
(c) A = 3, C = 0; AC = 0 Parabola
(d) A = 9, C = 9; AC = 81 Circle (A = C)
5
15. Sketch the graph of any two of the following:
(a)
x2
9
−
y2
25
=1
(b) (x − 2)2 + 16(y + 3)2 = 64
(c) y 2 + 2x − 4y − 4 = 0
Solution:
(a)
x2
9
2
y
− 25
=1
√
a=√
3, b = 5, c2 = a2 + b2 = 34, c = 34
5 < 34 < 6
√
center= (0, 0), vertices= (±3, 0), foci= (± 34, 0)
13
12
11
10
9
8
7
6
5
4
3
2
1
0
−1
−7−6−5−4−3−2−1 0 1 2 3 4 5 6 7
−2
−3
−4
−5
−6
−7
−8
−9
−10
−11
−12
−13
6
(b) (x − 2)2 + 16(y + 3)2 = 64
(x−2)2
(y+3)2
+
=1
64
4
√
a=√
8, b = 2, c2 = a2 − b2 = 60, c = 60
7 < 60 < 8
√
center= (2, −3), vertices= (2±8, −3), foci= (2± 60, −3)
2
1
0
−1 0 1 2 3 4 5 6 7 8 9 10 11 12
−8−7−6−5−4−3−2−1
−2
−3
−4
−5
−6
−7
(c) y 2 + 2x − 4y − 4 = 0
y 2 − 4y = −2x + 4
(y 2 − 4y + 4) = −2x + 4 + 4
(y − 2)2 = −2(x − 4)
4p = −2, p = − 12
focal diameter= |4p| = 2
vertex= (4, 2), focus= (4 − 21 , 2) = ( 27 , 2)
directrix: x = 4 + 12 = 92
4
3
2
1
0
−2
−1
0
−1
1
2
3
4
5
6
−2
−3
−4
7
16. Find the center, foci and vertices of either of the conics
below.
(a) 2x2 + y 2 − 8x − 6y − 1 = 0 (b) 2x2 − y 2 − 8x − 6y + 1 = 0
Solution:
(a) 2x2 + y 2 − 8x − 6y − 1 = 0
(2x2 − 8x) + (y 2 − 6y) = 1
2(x2 − 4x) + (y 2 − 6y) = 1
2(x2 − 4x + 4) + (y 2 − 6y + 9) = 1 + 8 + 9
2(x − 2)2 + (y − 3)2 = 18
2
(x−2)2
+ (y−3)
=1
9
18
2
Ellipse: c = a2 − b2 = 9, c = 3
√
Center= (2, 3), foci= (2, 3 ± 3), vertices= (2, 3 ± 18)
(b) 2x2 − y 2 − 8x − 6y + 1 = 0
(2x2 − 8x) − (y 2 + 6y) = −1
2(x2 − 4x) − (y 2 + 6y) = −1
2(x2 − 4x + 4) − (y 2 + 6y + 9) = −1 + 8 − 9
2(x − 2)2 − (y + 3)2 = −2
(y+3)2
(x−2)2
−
=1
2
1
√
2
Hyperbola: c = a2 + b2 = 3, c =√ 3
Center= (2, −3), foci=
√ (2, −3 ± 3)
vertices= (2, −3 ± 2)
8
17. Solve the system of equations









x + 2y − z = −3
2x − 4y + z = −7
−2x + 2y − 3z
=4
Solution:




−2E1 +E2 =E2 
→




x + 2y − z = −3
− 8y + 3z = −1
−2x + 2y − 3z
=4




2E1 +E3 =E3 
→








4E3 
→



x +
−




3E2 +E3 =E3 
→




− 81 E2
and
→
x + 2y − z = −3
− 8y + 3z = −1
6y − 5z = −2
2y −
z = −3
8y + 3z = −1
24y − 20z = −8
x + 2y −
− 8y +




1
− 11
E3 
z = −3
3z = −1
−11z = −11
x + 2y −
y −




z=1
y = 18 + 38 = 12
x = −3 − 2( 12 ) + 1 = −3
(−3, 12 , 1)
9
z = −3
= 81
z
=1
3
8z