<h2>
1:</h2>
- A mile unit is equal to 1760 yards. inches =mile * 63360 2*63360=126720
2 Miles = 126720 Inches
- There are 36 inches in a yard
3600/36= 100
3600 inches = 100 yards
6800/1000 = 6.8
6800 m = 6.8 km
15000/100=150
15000 cm=150 m
2:
a) Johnson ran yard in second= 9
1 minute = 60 second
Johnson ran yard in one minute= 9 *60= 540 yards
b) Johnson ran yard in second= 9
1 hour = 60 minutes *60 second= 1200 seconds
Johnson ran yard in one hour= 9 *1200= 10,800 yards
c) Johnson ran yard in one hour= 9 *1200= 10,800 yards
1 mile = 1760 yards
miles did he run in one hour= 10,800 / 1760 =6
Johnson ran 6 mile in one hour
The fourth class ends at 12:30 pm
<em><u>Solution:</u></em>
Given that Harold has 4 classes each morning
Each class is 1 hour long, and there are 10 minutes between classes
The first class is at 8 A.M
<em><u>To find: Time at which fourth class ends</u></em>
Since each is 1 hour long and 10 minutes gap between classes
First class = 8 A.M to 9 A.M
Second class = 9:10 A.M to 10 : 10 AM
Third class = 10 : 20 AM to 11 : 20 AM
Fourth class = 11 : 30 AM to 12 : 30 PM
Thus the fourth class ends at 12:30 pm
Answer:
first one is 32
the second one is 53
Step-by-step explanation:
i am certain that this is the answer
Answers:
33. Angle R is 68 degrees
35. The fraction 21/2 or the decimal 10.5
36. Triangle ACG
37. Segment AB
38. The values are x = 6; y = 2
40. The value of x is x = 29
41. C) 108 degrees
42. The value of x is x = 70
43. The segment WY is 24 units long
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Work Shown:
Problem 33)
RS = ST, means that the vertex angle is at angle S
Angle S = 44
Angle R = x, angle T = x are the base angles
R+S+T = 180
x+44+x = 180
2x+44 = 180
2x+44-44 = 180-44
2x = 136
2x/2 = 136/2
x = 68
So angle R is 68 degrees
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Problem 35)
Angle A = angle H
Angle B = angle I
Angle C = angle J
A = 97
B = 4x+4
C = J = 37
A+B+C = 180
97+4x+4+37 = 180
4x+138 = 180
4x+138-138 = 180-138
4x = 42
4x/4 = 42/4
x = 21/2
x = 10.5
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Problem 36)
GD is the median of triangle ACG. It stretches from the vertex G to point D. Point D is the midpoint of segment AC
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Problem 37)
Segment AB is an altitude of triangle ACG. It is perpendicular to line CG (extend out segment CG) and it goes through vertex A.
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Problem 38)
triangle LMN = triangle PQR
LM = PQ
MN = QR
LN = PR
Since LM = PQ, we can say 2x+3 = 5x-15. Let's solve for x
2x+3 = 5x-15
2x-5x = -15-3
-3x = -18
x = -18/(-3)
x = 6
Similarly, MN = QR, so 9 = 3y+3
Solve for y
9 = 3y+3
3y+3 = 9
3y+3-3 = 9-3
3y = 6
3y/3 = 6/3
y = 2
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Problem 40)
The remote interior angles (2x and 21) add up to the exterior angle (3x-8)
2x+21 = 3x-8
2x-3x = -8-21
-x = -29
x = 29
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Problem 41)
For any quadrilateral, the four angles always add to 360 degrees
J+K+L+M = 360
3x+45+2x+45 = 360
5x+90 = 360
5x+90-90 = 360-90
5x = 270
5x/5 = 270/5
x = 54
Use this to find L
L = 2x
L = 2*54
L = 108
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Problem 42)
The adjacent or consecutive angles are supplementary. They add to 180 degrees
K+N = 180
2x+40 = 180
2x+40-40 = 180-40
2x = 140
2x/2 = 140/2
x = 70
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Problem 43)
All sides of the rhombus are congruent, so WX = WZ.
Triangle WPZ is a right triangle (right angle at point P).
Use the pythagorean theorem to find PW
a^2+b^2 = c^2
(PW)^2+(PZ)^2 = (WZ)^2
(PW)^2+256 = 400
(PW)^2+256-256 = 400-256
(PW)^2 = 144
PW = sqrt(144)
PW = 12
WY = 2*PW
WY = 2*12
WY = 24
