PreCalc Chapter 7 Practice Test QandA - MathGuy.US · PreCalculus Chapter 7 Practice Test...
Transcript of PreCalc Chapter 7 Practice Test QandA - MathGuy.US · PreCalculus Chapter 7 Practice Test...
PreCalculusChapter7PracticeTest Name:____________________________
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Note the exponents of 2 in the equations. Any exponent greater than 1 will identify a system as
non‐linear. This system is non‐linear. Answer B.
Typically, you would begin solving a set of three equations by selecting pairs of equations and
eliminating the same variable in each pair. I’ll start this one by eliminating the variable .
Add 1st and 3rd equations Add 2nd and 3rd equations
1 9 3 3
2 4 2 2 12
We got lucky here because we were able to eliminate two variables at the same time by adding
the first and third equations. Normally, this would not happen, and you would have to solve the
set of two simultaneous equations which result. Let’s continue.
Solve for : Then, solve for : Then, solve for :
2 4 2 2 12 1
2 2 ∙ 2 2 12 2 4 1
4 2 12 2 1
2 8 3
4
Finally, test your results in one of the original equations, but not the one used to solve for .
Second equation: 2 4 3 9
Solution: , ,
Note: You can download a Microsoft Excel file from the following link
that will allow you to explore systems of 3 equations. Give it a try. http://www.mathguy.us/MathTidbits.php
PreCalculusChapter7PracticeTest Name:____________________________
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For problems 3 to 5, we need only write the form of the decomposition. We do not need to solve the
decomposition for the constants , and .
This rational function has no repeated linear factors in the denominator, so the decomposition is
straightforward:
2 36 6
This rational function has a repeated linear factor 7 , so the decomposition must include
each integral exponent of 7 , up to the exponent of the term in the rational function 2 :
3 15 7
This rational function has a quadratic function in the denominator so the decomposition must
take this into account:
2 52 4
PreCalculusChapter7PracticeTest Name:____________________________
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Write the form of the decomposition:
Multiply both sides by 1 5 : 15 39 5 1
Simplify: 15 39 5
Write the simultaneous equations and solve them:
15 5 39
Solve for : Then, solve for :
15 15
5 39 6 15
4 24
So, the partial fraction decomposition is:
15 391 5
PreCalculusChapter7PracticeTest Name:____________________________
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Write the form of the decomposition:
Multiply both sides by 3 : 7 36 3 3
Expand and simplify: 7 36 6 9 3
7 36 6 3 9
Write the simultaneous equations and solve them:
0 6 3 7 9 36
Solve for : Then, solve for : Then, solve for :
9 36 0 6 3 7
4 0 6 4 3 4 7
24 12 7
So, the partial fraction decomposition is:
7 363
PreCalculusChapter7PracticeTest Name:____________________________
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Write the form of the decomposition:
Multiply both sides by 1 1 : 10 2 1 1
Expand and simplify: 10 2
10 2
Write the simultaneous equations and solve them:
0 10 2
Now, let’s eliminate form the 2nd equation with some help from the 3rd equation.
10
2
2 12
Then,
Solve for : Then, solve for : Then, solve for :
2 12 0 2
4 0 4 2 0
3 12
So, the partial fraction decomposition is:
10 21 1
PreCalculusChapter7PracticeTest Name:____________________________
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Write the form of the decomposition:
Multiply both sides by 4 : 3 1 4
Expand and simplify: 3 1 4 4
3 1 4 4
Write the simultaneous equations and solve them:
0 1 4 3 4 1
Solve for : Then, solve for :
4 3 4 1
4 0 3 4 1 1
4 1
So, the partial fraction decomposition is:
3 14
PreCalculusChapter7PracticeTest Name:____________________________
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6 8 16
8 16 6
7 10 0
5 2 0
2, 5
When 2, we get:
2 6, so 4⇒ 2, 4 is a solution
When 5, we get:
5 6, so 1⇒ 5, 1 is a solution
So, our solutions are: , , ,
15 56
15 15 56
15 56
15 56 0
7 8 0
7, 8
When 7, we get:
15 7 8⇒ 7, 8 is a solution
When 8, we get:
15 8 7⇒ 8, 7 is a solution
So, our solutions are: , , ,
PreCalculusChapter7PracticeTest Name:____________________________
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2 66 41
Let’s use the Addition (i.e., Elimination) Method
2 66 multiply by 1 2 66
41 multiply by 1 41
25
5
When 5, we get:
5 41
25 41
16 ⇒ 4
5, 4 and 5, 4 are solutions
When 5, we get:
5 41
25 41
16 ⇒ 4
5, 4 and 5, 4 are solutions
So, the entire solution set is:
, , , , , , ,
Note: you can graph multiple conic sections (and lines) using the Algebra App, available at:
http://www.mathguy.us/Apps/AboutAlgebraMainApp.php.
On the opening page, click on the “Conic Sections” button in the “More Algebra” column.
On the left hand side of the Conic Sections page, click on the “Graph Multiple Equations”
button
You may enter up to 4 equations, using either General Form or Standard Form for each.
PreCalculusChapter7PracticeTest Name:____________________________
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29 4 17
Let’s use the Addition (i.e., Elimination) Method
29 multiply by 1 29
4 17 multiply by 1 4 17
4 12
4 12 0
2 6 0
2, 6
When 2, we get:
4 17
4 2 17
8 17
25 ⇒ 5
2, 5 and 2, 5 are solutions
When 6, we get:
4 17
4 6 17
24 17
7
Result: no real solutions when 6
So, the entire solution set is:
, , ,
PreCalculusChapter7PracticeTest Name:____________________________
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Let’s start by solving this for .
2 6
32
To graph this, do the following:
Graph the line: 3 .
The line will be solid because there
is an “equal sign" included in the
inequality.
Fill in the portion of the graph
above the line because of the
“greater than” portion of the
inequality.
To graph this inequality, do the following:
Graph the curve: 2 .
Some points on the curve:
2, 0.25 , 0, 1 , 2, 4
The curve will be solid because
there is an “equal sign" included in
the inequality.
Fill in the portion of the graph
below the curve because of the
“less than” portion of the
inequality.
PreCalculusChapter7PracticeTest Name:____________________________
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is the interior of a circle with
center , and radius √ .
To graph this inequality, do the following:
Graph the circle: 49. Some points on the curve:
0, 7 , 0, 7 , 7, 0 , 7, 0
The curve will be solid because there is an
“equal sign" included in the inequality.
Fill in the interior of the circle because of
the “less than” portion of the inequality.
(orange and green areas)
Graph the line: 4. The line will be dashed because there is no
“equal sign" included in the inequality.
Fill in the portion of the graph below the
line because of the “less than” portion of
the inequality.
(violet and green areas)
Graph the line: 2 3. The line will be dashed because there is no
“equal sign" included in the inequality.
Fill in the portion of the graph above the
line because of the “greater than” portion
of the inequality.
The green area is the area of intersection of the two linear inequalities.
PreCalculusChapter7PracticeTest Name:____________________________
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(orange and green areas)
Graph the parabola: .
Some points on the curve:
0, 0 , 2, 4 , 2, 4
The curve will be dashed because there is
no “equal sign" included in the inequality.
Fill in the portion of the graph above the
curve because of the “greater than”
portion of the inequality.
(violet and green areas)
Put this in slope intercept form
10 6 60 6 60 10
1053
Graph the line: 10 .
The line will be solid because there is an
“equal sign" included in the inequality.
Fill in the portion of the graph below the line because of the “less than” portion of the
inequality.
The green area (and contiguous magenta line) is the area of intersection of the two linear
inequalities.
PreCalculusChapter7PracticeTest Name:____________________________
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(orange and green areas)
Graph the circle: 49. Some points on the curve:
0, 7 , 0, 7 , 7, 0 , 7, 0
The curve will be solid because there is
an “equal sign" included in the
inequality.
Fill in the interior of the circle because of
the “less than” portion of the inequality.
(violet and green areas)
Put this in “ ” form
0
Graph the parabola: .
Some points on the curve:
0, 0 , 2, 4 , 2, 4
The curve will be dashed because there is no “equal sign" included in the inequality.
Fill in the portion of the graph above the curve because of the “greater than” portion of
the inequality.
The green area is the area of intersection of the two linear (and contiguous orange curve)
inequalities.
PreCalculusChapter7PracticeTest Name:____________________________
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Let’s use the Addition (i.e., Elimination) Method
5 20
2 7
3 27
3
Since the ship is in Q1, we must have positive and a
positive . So, 3. Then,
5 20
5 3 20
45 20
25 ⇒ 5 in Q1 So, , is the exact location
7 144
Let’s use the Substitution Method
7 7 144
7 144
7 144 0
16 9 0
16, 9
When 16, we get:
7 16 9⇒ 16, 9 is a solution
When 9, we get:
7 9 16⇒ 9, 16 is a solution
So, the two numbers are: and
PreCalculusChapter7PracticeTest Name:____________________________
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37 5
Let’s use the Substitution Method
5 5 37
10 25 37
2 10 12 0
5 6 0
6 1 0
6, 1
When 6, we get:
6 5 1⇒ 6, 1 is a solution
When 1, we get:
1 5 6⇒ 1, 6 is a solution
So, the two numbers are: and
2 2 42 90 with the dimensions of the rectangle being: by
Let’s use the Substitution Method
21 21 90
21 90
21 90 0
15 6 0
6, 15
When 6, we get:
21 6 15⇒ 6, 15 is a solution
When 15, we get:
21 15 6⇒ 15, 6 is a solution
So, the dimensions are: ft. by ft.