1. Classify each angle as acute, right, obtuse, or straight
2. Find the perimeter or circumference, as indicated.
a.A rectangle measures 10 inches by 5 inches. Find its perimeter.
b.A square has a side that measure 6 centimeters. Find its perimeter
c.John planted a garden that is 5 feet by 3 feet. How much fencing is needed to fence in theperimeter?
d. Find the circumference of a circle with a radius of 6 inches.
e. Find the circumference of a circle with a diameter of 18 centimeters.
3. Find the area or volume, as indicated.
a. Find the area of a rectangular carpet that measures 10 feet by 8 feet.
b. Find the area of a triangle with a base of 7 centimeters and a height of 11 centimeters.
c. Find the area of a circle with a radius of 9 inches.
d. Find the volume of a rectangular box that measures 6 inches by 4 inches by 2 inches.
e. Find the volume of a sphere with a radius of 8 meters.
College Mathematics C17V
Mathematics
- 1. a. 35° = acute angle
b. 135° = obtuse angle
c. 90° = right angle
d. 99° = obtuse angle
e. 180° = straight angle
2. a. Perimeter of a rectangle = 2 • length + 2 • width
= 2 • 10 + 2 • 5
= 30
The perimeter is 30 inches.
b. Perimeter of a square = 4 • s
= 4 • 6
= 24
The perimeter of a square is 24 centimeters.
c. To find the amount of fencing needed, we find the distance around, or perimeter. The formula for the perimeter of a rectangle is P = 2 • length + 2 • width. We use this formula and replace length and width by 5 feet and 3 feet.
Perimeter of a rectangle = 2 • length + 2 • width
= 2 • 5 + 2 • 3
= 16
The amount of fencing needed is 16 feet.
d. Circumference of a circle = 2 • π • r
= 2 • 3.14 • 6
= 37.68
The circumference of a circle is 37.68 inches.
e. Circumference of a circle = π • d
= 3.14 • 18
= 56.52
The circumference of a circle is 56.52 centimeters.
3. a. Area of a rectangle = length • width
= 10 • 8
= 80
The area is 80 feet.
b. Area of a triangle = 1/2• base • height
= 1/2• 7 • 11
= 38.5
The area is 38.5 centimeters.
c. Area of a circle = π • (radius)2
= 3.14 • (9)2
= 254.34
The area is 254.34 inches.
d. Volume of a rectangle = length • width • height
= 6 • 4 • 2
= 48
The volume of a rectangular box is 48 inches.
e. Volume of a sphere = 4/3• π • (radius)3
= 4/3• 3.14 • (8)3
= 2,143.5733
The volume of a sphere is 2,143.5733 meters.
4. a. These triangles are not congruent because they do not have the same three sides.
b. These triangles are congruent because they have the same Angle-Side-Angle(ASA).
5. a. Since, the triangles are similar, corresponding sides are in proportion. Thus, the ratio of n to 7 is the same as the ratio of 8 to 4, or
n/7 = 8/4
To find the unknown length n, we set cross products equal.
n/7 = 8/4
n • 4 = 7 • 8
n = 56/4
n = 14
The missing length is 14 units.
b. Since the triangles are similar, corresponding sides are in proportion. Thus, the ratio of 7.5 to 5 is the same as the ratio of n to 8, or
7.5/5 = n/8
To find the unknown length n, we set cross products equal.
7.5/5 = n/8
5 • n = 7.5 • 8
n = 60/5
n = 12
The missing length is 12 units.
6. a. Joe had the highest sales for the year 2010.
b. Joe sold approximately 1.8 million dollars more than Julie.
7. a. pop
b. 4%
8. a. mean = sum of items/number of items
= 3.8+3.1+2.6+3.5+3.0+4.0+1.6+2.6+2.1/9
= 2.9222
median = 3.0
mode = 2.6
b. mean = 20,000+35,000+20,000+19,000+26,000+35,000+42,000/7
= 28,142.857
median = 26,000
mode = 20,000 and 35,000
9. To calculate the grade point average, we need to know the point values for the different possible grades. The point values of grades commonly used in colleges are given below:
A: 4, B: 3, C: 2, D: 1, F: 0
Now, to find the grade point average, we multiply the number of credit hours for each course by the point value of each grade. The grade point average is the sum of these products divided by the sum of the credit hours.
Course
|
Grade
|
Point Value of Grade
|
Credit Hours
|
Point Value × Credit Hours
|
Math
|
A
|
4
|
4
|
16
|
Science
|
A
|
4
|
4
|
16
|
English
|
C
|
2
|
3
|
6
|
Physical Education (PE)
|
B
|
3
|
1
|
3
|
History
|
B
|
3
|
3
|
9
|
Totals:
|
15
|
50
|
grade point average = 50/15 = 3.33
The student earned a grade point average of 3.33.
10. a. 0
b. 3/6 = 1/2
c. 4/6 = 2/3
d. 2/6 = 1/3
e. 2/6 = 1/3
11. a. +200 feet
b. -25°
12. a. -2 < 5
b. 15 > 10
c. -4 > -8
d. 25 > -96
e. -1.45 < 1.5
13. a. 4
b. 4
c. -5
d. 6
e. -2
14. a. -37
b. -9/2 + 19/8
= -36+19/8
= -17/8
c. 10.32
d. -5.8
e. 42 - 52/56
= -10/56
= -5/28
15. a. -24
b. -6/7
c. 8
d. 32.4
e. 3/8 • -2/1 = -¾
16. a. 5x -2 =23
5x = 25
x = 5
b. 4y - 16 = 0
4y = 16
y = 4
c. 5 - x = -15
-x = -15 - 5
x = 20
d. 1.5x + 1 = 5.5
1.5x = 4.5
x = 3
e. -3x + 9 = 27
-3x = 18
x = -6
17. a. 3y + 24 = y - 12
3y - y = -12 - 24
2y = -36
y = -18
b. 3y - 7 = -3y + - 19
3y + 3y = 7 - 19
6y = -12
y = -2
c. 15x - 8x + 7 = -42
7x = -49
x = -7
d. -x + 5x + 6 = 8x + 36
4x - 8x = 36 - 6
-4x = 30
x = -15/2 or -7.5
e. -3(x + 5) = -18
-3x -15 = -18
-3x = -3
x = 1
18. a. x + 7
b. 20 - x
c. 50 - x
d. 2x + 4
e. x • 8
19. a. 2x + 18 = 22
2x = 22 - 18
2x = 4
x = 2
b. 3x • 7 = 63
3x = 63/7
3x = 9
x = 3
20. a. Let the Doug's age be x
Doug's age = 2x
Sean's age = 2x + 5
Now, 2x + (2x + 5) = 29
4x = 29 - 5
4x = 24
x = 6
Doug's age = 2x = 2 • 6 = 12
Sean's age = 2x + 5 = 2 • 6 + 5 = 17
b. Let the number of score of Larry be x
2x + x = 48
3x = 48
x = 16
The number of score of Larry is x = 16
References
Elayn Martin-Gay (2011). Basic College Mathematics (4th Ed.). Boston: Pearson.
