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MATH 1630 Internet Spring, 2009 Quiz 3 Name: KEY (78 Possible Points)
This quiz includes sections 2.1, 2.2, 2.3, 2.4, 2.5, 3.1 and 3.
You must show all work to receive full credit. Points will be deducted for work that is not neat or orderly. Do all work on these sheets. Round all decimal answers to the nearest one-hundredth.
- (6 points) Solve the following equation: 5 3 4 x 4 x 2
Weโll eliminate the fractions by multiplying both sides by the common denominator (x+4)(x-2)
Use the distributive property to multiply and then cancel. 4
x x x x x x
x
+ โ โ^ โ โ= + โ
x x
โ ( x + 4 ) ( x โ 2 )^3
x โ 2
2 2 2 2 2 2 2
I'm switching the two sides so everything will end up on the left. 4 8 10 2 4 8 2 10 2 2 4 6 10 0 2 2 3 5 0 2
x x
x x x x x x x x x x x x x x
x x x x x x x x x x x x
( 2 5 )( 1 )^ 0 Set each factor equal to 0 and solve.
There are two possible solutions: 5 and 1. 2 You should plug both solutions into the original equa
x x x x x x x x x x
x x
tion to ensure they each make the equation a true statement.
- (4 points) Solve the following equation: 5 t^2 + 5 t + 3 = 0
Weโll use the quadratic formula to solve:
a = 5, b = 5, c = 3
Since the 35 is not a real number, then we conclude there are NO REAL SOLUTIONS.
x b^ b^ ac a
x
โ ยฑ โ โ^ ยฑ^ โ โ ยฑ โ
=โ^ ยฑ^ โ
- (6 points) Showing all appropriate calculations, determine the domain and range of
f ( x ) = โ 2 x^2 + 5 x โ 7
Domain:
The domain (the set of all possible value of x) is the set of Real Numbers. There are no restrictions on x.
Range:
Range is the set of all possible outputs (values of f(x) or y). The graph of the equation is a parabola pointing down. The vertex, at the top of the parabola, gives the maximum value of f(x).
x b a f
= โ^ = โ^ = โ =
Notice the value of โaโ is -2. This means that the parabola points down. The vertex is at the top of the parabola (i.e. the vertex gives us the maximum value of y.)
The range is any real number less than or equal to 3.875.
- A ball thrown into the air vertically has its height above the ground determined by
h t ( ) = โ 16 t^2 + 104 t , where t is in seconds and h(t) is in feet. Calculate each of the following quantities.
Show all necessary work. (5 points each)
a. How many seconds until the ball hits the ground.
When the ball hits the ground, the height (h(t)) will equal 0. So we need to solve the equation:
We can solve this by factoring: 16 104 0 0 16 104 0 16 104 104 0 104 16 104 16 104 16 16
t t
t t t t t t t t
There are two times when the ball is on the ground. The first (t = 0 seconds) is when we are getting ready to throw the ball. The second (t = 6.5 seconds) indicates that the ball hits the ground after 6. seconds.
b. Calculate the maximum height the ball reaches.
The graph of the equation for the height of the ball is a parabola point down. The vertex gives the maximum height of the ball and the number of seconds for the ball to reach that maximum height.
x b a
h
= โ^ = โ^ = โ =
The ball reaches a maximum height of 169 feet after 3.25 seconds.
c. Calculate the time(s) the ball is at a height of 100 feet. We need to solve the equation
2 2 2
2 2
Use the quadratic formula: a = -16, b = 104, c = -100. 4 104 104 4 16 100 104 10816 6400 104 4416 2 2 16 32 32 104 4416 32 104 4416
t t t t t t
t b^ b^ ac a
t
t
โ ยฑ โ โ^ ยฑ^ โ^ โ^ โ โ ยฑ โ โ ยฑ
= โ^ ยฑ
= โ^ + 1.
t
= โ^ โ =
The ball reaches a height of 100 feet twice. The first is at 1.1733 seconds (on the way up), and the second time at 5.3267 seconds (on the way down).
7. The profit function for a firm making widgets is P x ( ) = 88 x โ x^2 โ 1200 , where x is the number of
widgets. Calculate each of the following quantities. You must show all work.
a. (5 points) Calculate the break-even point(s).
The break-even point is defined as the number of goods to be made and sold so that Profit = 0. We need to solve the equation
2 2
2 2
Use the quadratic formula: a = -1, b = 88, c = - 4 88 88 4 1 1200 88 7744 4800 88 2944 2 2 1 2 2 (^88 2944) 16.8707 17 2 (^88 2944) 71.1293 71 2
x x x x
x b^ b^ ac a
x
x
โ ยฑ โ โ^ ยฑ^ โ^ โ^ โ โ ยฑ โ โ ยฑ
= โ^ + = โ
= โ^ โ = โ
There are two break-even points.
b. (5 points) Calculate the maximum possible profit and the corresponding number of widgets to be made and sold.
We need to determine the coordinates of the vertex.
We maximize profit with the production and sale of 44 widgets. (44) 44 88 44 1200 1936 3872 1200 736 The maximum profit is $736.
x b a
P
= โ^ = โ = =
- Use the following matrices for problem 8
3 1 5 1 2 1 4 , , 5 2 3 3 4 2 3
A B C
โก^ โ โ^ โค โก โ^ โค โก โ โค
โฃ โฆ โฃ โ^ โฆ โฃ โโฆ
a. (2 points each) List the dimensions of each matrix.
A: 2 by 2 (2 rows, 2 columns) B: 2 x 3 (2 rows, 3 columns) C: 2 by 2 (2 rows, 2 columns)
b. (2 points) In matrix B, list the coordinates of the entry โ-3โ. 2, 2 (2nd^ row, 2nd^ column)
c. (2 points) List the dimensions of the matrix BT. _________________
B T represents B-transpose (i.e. switch the rows and columns). If matrix B is 2 x 3, then matrix B- transpose is 3 x 2.
d. (4 points) Showing all work, determine 2A โ 3C
2 3 2 3 3 1 3* 2 1* 2 6 2 2 2 5 2 5* 2 2* 2 10 4 1 4 1* 3 4* 3 3 12 3 3 2 3 2* 3 3* 3 6 9 6 2 3 12 3 14 2 3 10 4 6 9 4 13
A C A C
A
C
A C
โก โ^ โ^ โค โก โ^ โ^ โค โกโ โ โค
โก โ^ โค โกโ โ^ โ^ โค โก โ โค
โฃ โ^ โฆ โฃ โ^ โ^ โ^ โฆ โฃ โ โฆ
โก โ^ โ^ โค โก โ^ โค โก โ^ โ โค
