Download Answers: MAT 110 Practice Test (Chapter 7)-PDF

Math 110 Chapter 7 Practice Test Answers 1. Let f(x) = ‐5x + 3. Find f (‐3). a. F(‐3) = ‐5(‐3) + 3= 15 + 3 = 18 2. Find the slope of the line through the given points. (‐3, 0) and (3, 18). a. M=
m= = = = 3 OR m = = = 3 3. Determine if the two given lines are parallel. Answer yes or no. 3y = 2x – 2 8x – 12y = ‐2 a. For 2 lines to be considered parallel, they must have the same slope, but not be the same line. To find the slope… Y = mx + b i. 3y = 2x – 2 Y= x – 2 M= ii. 8x – 12y = ‐2 ‐12y = ‐8x ‐ 2 Y = x ‐ Y= x + M= b. The slopes are the same, and the two equations are different; yes, they are parallel. 4. Find and equation of the line passing through the point (‐2, ‐4) with slope = 3. Write in slope‐intercept form. m = 3, x = ‐2, y = ‐4 a. Y – y1 = m(x ‐ x1) Y – (‐4) = 3(x – (‐ 2) Y + 4 = 3x + 6 Y = 3x + 2 5. Find and equation of the line passing through (‐4, ‐2) and (‐2, 4). Write in slope‐ intercept form. a. M = = = = = 3 Y‐y1 = m (x‐x1) y – (‐2) = 3(x – (‐4)) OR y – 4 = 3(x – (‐2)) y + 2 = 3(x + 4) y – 4 = 3(x + 2) y + 2 = 3x + 12 y – 4 = 3x + 6 y = 3x + 10 y = 3x + 10 6. & 7. Solve the following system of equations by substitution. 2x + 2y = 4 y = ‐x 2x + 2 (‐x) = 4 2x ‐2x = 4 NO SOLUTION 8 and 9. Solve the system of equations using the addition method. 4x – 3y = 1 3x – 4y = 4 Methods will vary; ‐3 (4x – 3y = 1) 4 (3x – 4y = 4) ‐12x + 9y = ‐3 12x – 16y = 16 ‐7y = 13 Y =‐ 4x – 3(‐13/7) = 1 4x + 39/7 = 7/7 4x = ‐32/7 x= ‐32/28 x = ‐8/7 10 and 11. Solve the following system of equations by graphing. Y = 3x – 7 y + 2x = 8 15
10
5
y=3x‐7
0
‐4
‐2
0
2
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y=‐2x+8
‐5
‐10
‐15
12. The sum of two integers is 49. Find the two numbers if the larger is 8 less than twice the smaller. Let x be the smaller and y be the larger of the two integers. X + y = 49 2x – 8 = y Using substitution x + (2x – 8) = 49 x + 2x – 8 = 49 3x – 8 = 49 3x = 57 x = 19 Now, 19 + y = 49 y = 30 y‐ Intercept (when x = 0): 0 + 4y = 8 y = 2 y‐int = (0, 2) x‐ Intercept (when y = 0): 2x + 0 = 8 x = 4 x‐int = (4, 0) 13. Graph by intercepts: 2x + 4y = 8 4y = ‐2x + 8 Y = ‐ x + 2 y= ‐2x + 8
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12
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8
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y= ‐2x + 8
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2
0
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‐2 0
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6
14. Graph by using slope and y intercept. 3x + 6y = 12. Y intercept (when x=0): 6y = 12, y = 2 (0, 2) Slope (m) = (find y = mx + b form first) 6y = ‐3x +12 Y = ‐ x + 2 m = ‐ Y = ‐(1/2)x + 2
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3
2
Y = ‐.5x + 2
1
0
‐4
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0
2
4
15. Graph by plotting points. 2x + y = 6. Y = ‐2x + 6 X ‐3 ‐2 ‐1 0 1 2 3 Y 12 10 8 6 4 2 0 Y = ‐2x + 6
15
10
Y = ‐2x + 6
5
0
‐5
0
5
16. Graph 3x + 4y ≤ 12. 17. Graph x=1 x=1
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x=1
0
‐50
0
0.5
1
1.5
‐100
18. Draw the graph of a line whose slope is zero. m=0
1.2
1
0.8
0.6
0.4
0.2
0
‐200
‐100
m=0
0
100
200
19. Graph y > 2x 20. Graph x = y x = y
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2
0
‐4
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21. Graph the system: 3x + 2y ≥ 12 x < y 2y ≥ ‐3x + 12 Y ≥ ‐ + 6 22. At Rhonda’s diner, three loaded baked potatoes and five cheeseburgers provide 3820 calories. One loaded baked potato and four cheeseburgers provide 2370. How many calories in one baked potato? How many calories in one cheeseburger? Let b = the number of calories in a baked potato, and c = the number of calories in a cheeseburger. ** This is only one way to do this problem. You can multiply either equation. 3b + 5c = 3820 b+ 4c = 2370 3b + 5c = 3820 ‐3(b + 4c = 2370) 3b + 5c = 3820 ‐3b – 12c = ‐7110 Using the addition method, we have the equation; ‐7c = ‐ 3290 c= 470 Now, using substitution, b+ 4(470) = 2370 b+ 1880 = 2370 b= 490 23. I look out my front window and see children and dogs playing in my front yard. When I look out, I see 18 heads and I count 60 feet and paws. How many children and how many dogs are playing in my front yard? Dogs have 4 feet, children have 2. Let x be children and y be dogs. There are 18 heads. Each child and dog has 1 head x+ y = 18 ‐2(x+ y = 18) ‐2x ‐2y = ‐36 2x+ 4y = 60 2x+ 4y = 60 2x + 4y = 60 2y = 24 Y= 12 x+ y =18 x = 6 Check: dogs paws + children’s feet =60 : 12*4 + 6*2 = 48 + 12 = 60 24. using the slope‐intercept form, write a system of equations that are perpendicular. Answers will vary; as long as your form is y1 = mx + b and y2 = ‐ (1/m) x + b, your equations should be correct. 25. Write the equation of the line through (2, ‐7) which is perpendicular to 5x + 2y = 18. First, let’s put the equation into y = form. 2y = ‐5x + 18 y = ‐ (5/2) x + 9 y– y1 = m(x – x1) y– (‐7) = m(x – 2) What is m? Since we’re looking for perpendicular lines, the slope of our new line is the negative reciprocal of the slope of our original. y‐ (‐7) = (2/5) (x – 2) y+ 7 = (2/5) x ‐ 4/5 y= (2/5) x + (‐35/5 ‐ 4/5) y= 2/5x – 39/5 7 = 35/5