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1. Solve 15x = 2x2 + 16
Put everything on one side of the equation, then use the quadratic formula:
15x = 2x2 + 16
0 = 2x2 – 15x + 16
x = 15 ±√(152 – 4(2)(16))
2(2)
x = 15 ±√97
4
x = 1.29, 6.21
2. The width is 7 feet less than the length; the area is 18 square feet. Find the length and
width.
Let x be the length of the flowerbed. Then, x – 7 would be the width. Since, area is equal to
length x width, we have the following equation:
(x)(x – 7) = 18
Multiply, put everything on one side of the equation, then factor to solve for x:
(x)(x – 7) = 18
x2 - 7x = 18
x2 - 7x -18 = 0
(x – 9)(x + 2) = 0
x = 9, -2
The only possible answer for x is x = 9 (a length can’t be negative). Therefore, the sides are 9
and 2.
3. Solve x2 + 4x – 12 < 0
x2 + 4x – 12 < 0
(x + 6)(x – 2) < 0
This happens when x is between -6 and 2. In solution set notation, this is {x | -6 < x < 2}; in
interval notation, this is (-6, 2).
4. Solve (x + 3)/(x – 4) < 0
(x + 3)/(x – 4) < 0
This happens when x is between -3 and 4. In solution set notation, this is {x | -3 < x < 4}; in
interval notation, this is (-3, 4).
5. Find f(-13) for f(x) = 3√(2x – 1)
Just plug -13 in for x:
f(x) = 3√(2x – 1)
f(-13) = 3√(2(-13) – 1)
f(-13) = 3√(-24)
f(-13) = -2.884499
6. Rewrite (3√(7xy))4
(3√(7xy))4 = ((7xy)1/3)4 = (7xy)4/3
7. Simplify 5√(p14q9r23)
√(p14q9r23) = (p14q9r23)1/5 = (p14/5q9/5r23/5)
5
8. Solve 3√(3y + 6) + 2 = 5
√(3y + 6) + 2 = 5
√(3y + 6) = 3
3y + 6 = 27
3y = 21
y=7
3
3
9. Solve for t: L = (Mt + g)/t
L = (Mt + g)/t
Lt = Mt + g
Lt – Mt = g
t(L – M) = g
t = g/(L – M)
10. Graph f(x) = |x| - 4
x
f(x)
5
-4
-3
-2
-1
0
1
2
3
4
5
1
0
-1
-2
-3
-4
-3
-2
-1
0
1
2
1
0
5
-4
-3
-2
-1
0
1
2
3
4
5
-2
-1
0
1
2
3
4
5
-1
-2
-3
-4
-5
11. Graph f(x) = x2 – 1
x
f(x)
5
-4
-3
-2
-1
0
1
2
3
4
5
24
15
8
3
0
-1
0
3
8
15
24
30
25
20
15
10
5
0
5
-4
-3
-5
12. Simplify (x3y/pq2)-3
(x3y/pq2)-3
x-9y-3/p-3q-6
p3q6/ x9y3
(pq2/ x3y)3
13. Subtract and simplify 4/(5a2 – 5a) – 2/(5a – 5)
4
5a – 5a
-
2
5a - 5
4
a(5a – 5)
2
5a - 5
4
a(5a – 5)
2a
a(5a – 5)
2
4 - 2a
a(5a – 5)
14. Find the domain of f(x) = 15/(3x – 8)
The domain is all real numbers except those that make the denominator equal to 0.
3x – 8 = 0
3x = 8
x = 8/3
The domain is all real numbers except x = 8/3. This can be written as (-∞, 8/3) U (8/3, ∞).
15.
a) The two points we have is (9, 74) and (23, 92). We can use these points to find the slope, then
to find the y-intercept:
slope = (92 – 74)/(23 – 9) = 18/14 = 9/7
74 = (9/7)(9) + b
74 = 11.57 + b
b = 62.42857
The line is y = (9/7)x + 62.43.
b) Plug x = 18 into the equation:
y = (9/7)x + 62.43
y = (9/7)(18) + 62.43
y = 85.57
16.
Let x be the original cost of the meal (before the tip). Then, we can make the equation:
$30.50 = x + 0.15x
$30.15 = 1.15x
17.
An odd integer can be described as 2n + 1 for any integer value of n. The next odd number after
that one is 2n + 3. Therefore the problem can be reduced to the following equation:
255 = (2n + 1)(2n + 3)
18.
Let x be the number of $5 bills and y be the number of $1 bills. We can make 2 equations:
x + y = 22
5x + y = 50
19. Solve 3(r – 6) + 2 > 4(r + 2) - 21
3(r – 6) + 2 > 4(r + 2) - 21
3r – 18 + 2 > 4r + 8 - 21
3r – 16 > 4r – 13
-3 > r
As a graph, this looks like:
○
-5
-4
-3
-2
-1
0
1
In set notation, this is {r | r < -3}
In interval notation, this is (-∞, -3).
20. Graph 2x + 3y ≥ 6
2x + 3y ≥ 6
3y ≥ -2x + 6
y ≥ (-2/3)x + 2
Graph the line y = (-2/3)x + 2, then shade above it. The actual line should be solid, not dotted,
indicating that the line is part of the solution.