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1 Simplify : [5 - 2/x]/[1/5 + 6/x]
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a [25 - 10x]/[1 + 30x]
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| b [25x - 10]/[1 + 30x] |
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c [25x - 10]/[x + 30]
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d [25 - 10x]/[x + 30]
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e nota
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2. Simplify : [2/x2
- 1/xy - 1/y2]/[1/x2
- 3/xy + 2/y2] (Hint:
Factor and cancel.)
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a [2y - x]/[y - 2x]
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b [2y + x]/[y - 2x]
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c [2y - x]/[y + 2x]
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d nota
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3. Simplify : [2 -
1/x - 1/x2]/[1
- 3/x + 2/x2] (Hint: Factor
and cancel.)
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| a [2x + 1]/[x + 2]
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b [2x + 1]/[x - 2] |
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c [2x - 1]/[x - 2] |
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d nota |
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6.6
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4. Solve: 3/(x + 1) = 5/(x -1). (Hint:
cross-multiply.) The answer for x is a value in the interval
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| a (5,10) |
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b (0, 5) |
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c (-5, 0) |
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d nota
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| 5. Solve 6/x
- x/3 = 1 (Hint: Multiply both sides by the
LCD.) The values for the solution for x are in the intervals
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| a (-7, 0) or
(0,4) |
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b (0, 7) or
(-4,0) |
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c (-3, 0) or
(0,2) |
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d nota |
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e nota
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6. Clean-up costs for pollution
removal in a river are given by f(x) = 250x/(100 - x), in millions of
dollars.
What is the percentage x cleaned if f(x) = 375 (million dollars)?
The value for x is in the interval:
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| a (0,30) |
| b (30,50) |
| c (50,70) |
| d (70,90) |
| e nota
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| 6.7
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7.
Solve for x in the formula: A = xy/(x + y)
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a (Ax - Ay)/y |
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b (Ax +Ay)/y |
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c Ay/(y + A) |
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d Ay/(y - A) |
| e nota |
| 8. A pool can be filled by one pipe
in 3 hours and by a second pipe in 6 hours. How long will it take using
both pipes to fill the pool? (Hint: See Example 4) |
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a 1 hour |
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b 2 hours |
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c 5/2 hours |
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d nota |
| 6.8 |
9. Hint: See Example 1. Also read
the purple boxes on pages 440 and 441. The cost C of an airplane varies directly
as the number of miles M in the trip. Yo, a 3000 -mile trip costs
$400. What is the cost of a 450-mile trip? Note this direct variation
can also be thought of as a proportion problem:
(3000 miles)/($400) = (450 miles)/x. Thus, you can cross-multiply
and solve for the cost x, i.e. solve: 3000/400 =
450/x. However, it would not hurt to read Example 1. You can use exactly
the same method on this one. Note : This problem is exactly like
application exercise # 11 on page 448. |
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a $ 20
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b $ 40 |
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c $ 60 |
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d nota
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10. Hint: See Example 3. Also read
the purple box on page 443. The current I in a circuit varies inversely
as the resistance R. The current I is 20 amperes when the resistance R
is 5 ohms. What is the current for a resistance of 16 ohms. Note
this inverse variation problem can also be thought of
as a proportion
problem:
(20 amperes)(5 ohms) = I ·(16 ohms) .
Thus, you can solve for I easily by dividing both sides by 16. However,
it would not hurt to read Example 3. You can use exactly the same method
on this one. Note this problem is almost like application exercises # 19
and # 20 on pp 448-449. |
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a 5.25 amperes |
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b 6.25 amperes |
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c 7.25 amperes |
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d nota |
| 7.1 |
11. Simplify these two
problems:
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a 4x and 2x2 |
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b 4x2 and 2x2 |
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c 4x2 and 2x |
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d nota |
12. Simplify this problem:
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a -4x3 |
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b -4x2 |
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c 4x2 |
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d undefined because you cannot have the cube root of a negative
number. |
| e nota |
| 13. Extra Credit. 7.2.
Simplify 82/3 = 4 |
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a 2 |
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b 4 |
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c 6 |
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d 8 |
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e nota
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