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Math 140 Placement Verification Test Solution Review of Similar Type Problems

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Math 140. Placement Verification Test. Solution Review of. Similar Type Problems. Problem 1 ... Problem 12. Solve this equation. ... Problem 16 ... – PowerPoint PPT presentation

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Title: Math 140 Placement Verification Test Solution Review of Similar Type Problems

1
Math 140 Placement Verification Test Solution
Review of Similar Type Problems
2
Problem 1
• Simplify 12-1 3(4-2) .
• 12-1 4-2

3
Problem 2
• Simplify (3x-6)(-x4)3 .
• 38(9x)-2

3x-64?3(-1)3(3-82?2 x2)
-31-4 x62 -x 8/27
4
Problem 3

Note 625 54 (-1) (-1)3.
-54/3 51/3 -551/3 51/3
-451/3
5
Problem 4
• Perform these operations and simplify (3x -
4)2 - (x - 3)(8x 1).

(9x2 -2 3 4x 16) - (8x2 (1- 3 8)x - 3)
(9-8)x2 (-2423)x 16 3
x2 - x 19
6
Problem 5
• Simplify
.

7
Problem 6
• Perform these operations and simplify

8
Problem 7
• Perform the indicated operations and simplify
• (x - 2b2)3 .

(x - 2b2)3 x3 - 3(2b2) x2 3(2b2)2 x -
(2b2)3 x3 - 6b2 x2 12b4 x -
8b6
9
Problem 8
• Simplify __________
• 25

__________ 25
__________ 5
10
Problem 9
• Solve 7 - 5x lt 2(x -21).
• 7 - 5x lt 2x - 42
• 7x lt - 49
• x gt - 49/(-7)
• x gt 7

11
Problem 10
• Solve for x in the equation

12
Problem 10 Continued
• Solve for x in the equation

Alternate approach Multiply by LCD acx. Then,
cx 2acx ac ax 0
(c 2ac a)x ac
13
Problem 11
• Solve Solve 3 - 2x lt 5.

a) 3 - 2x lt 5 -2x lt 2 x gt -1
or b) (3 - 2x) lt 5 2x lt
8 x lt 4
Thus, -1lt x lt 4.
14
Problem 12
• Solve this equation. Write the sum of the
answers that is, write the result when all
- 1) 16.

x2 (5-1)x - 5 16 x2 4x - 21 0 (x - 3)(x
7) 0 x 3 0 or x 7 0 x 3, -7 is
solution set. Sum is 4.
Or from quadratic formula x -b (b2
4ac)1/2/(2a). Sum is 2b/(2a) -4.
15
Problem 13
• Perform the indicated operations
• on the expression (16/81)-3/4.

(81/16)3/4 (34/24)3/4 (3/2)43/4
(3/2)43/4 (3/2)3 27/8
16
Problem 14
• Perform the indicated operations
• on the expression .

17
Problem 15
• Assuming a gt 0 and b gt 0, simplify the indicated
expression equivalently so that all exponents are
positive (3ab4)-3 .

• (9ab6)-2

Note two negative exponents gt
flipping fraction avoids later trouble
with signs.
322a2b62 33a3b43
3/a
18
Problem 16
• Assuming a gt 0 and b gt 0, simplify the indicated
expression equivalently so that all exponents are
positive (144a-2/3b-4/3)3/2 .

1223/2a(-2/3)(3/2)b(-4/3)(3/2)
123a-1b-2 1728/(ab2)
19
Problem 17
• Simplify to standard form by performing the
indicated operations on the expression
• (x 1)(3x - 4)2 - (x - 3)(x 3)(8x
1).

(x 1)(9x2 24x 16) - (x2 - 9)(8x
1) 9x3 (-249) x2 (16-24)x 16 (8x3 x2
-72x - 9) x3 - 16x2
64x 25
20
Problem 18
• Factor the expression over the integers or state
it is prime if it cannot be factored x2 - 6x
- 16

Work systematically. There are 5 possibilities
Two with 1 16 gt (x - 1) (x 16) (x 1)
(x - 16), Two with 2 8 gt (x - 2) (x 8) (x
2) (x - 8), One with 4 4 gt (x - 4) (x 4)
(x 4) (x - 4).

But to get the first
degree middle term (- 6x ) only the combination
of factors (x 2) (x - 8) works.
x2 - 6x - 16 (x 2) (x - 8).
21
Problem 19
• Simplify by rationalizing the denominator
• of the expression 25 .

__________ 22 (51/2)2
22
Problem 20
• Simplify in factored form

23
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