Answer : option C
We analyze the attached graph
First, Ricky walked away from mall at the constant speed for 30 seconds
Second, he stopped walking for 40 seconds
Third, he walked towards mall at a constant speed for 10 seconds
Fourth , both walked away for next 20 seconds
So total walk time = 30 + 40 + 10 + 20= 100 seconds
Total time is 100 seconds. so we cannot choose option B and d because the graph ends before 100 seconds.
In option A, Ricky walked away from the mall after 70 seconds. so we cannot choose option A.
Option C is the graph best represents Ricky's motion.
Answer:
x = 6
Step-by-step explanation:
5 (x - 3) + 2 = 5 (2x - 8) - 3
<em>remove the parenteses</em>
5x - 15 + 2 = 10x - 40 - 3
<h3><em>calculate</em></h3>
5x - 13 = 10x - 43
<h3><em>move the terms</em></h3>
5x - 10x = -43 + 13
<h3><em>collect like terms </em></h3><h3><em>calculate</em></h3>
-5x = -30
<h3><em>divide both sides</em></h3><h3><em /></h3>
Solution
x = 6
Answer:
h+5(h+4)=40
Brian: 3 1/3 hours
Annie: 39 2/3 hours
Step-by-step explanation:
h+5(h+4)=40
Solve the equation
h+5h+20=40
6h=20
h=10/3 or 3 1/3
Annie = 5(10/3 + 4)= 39 2/3
Answer:
24.00
You have to add all 4 sides of the figure.
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
------------------------------------------------------
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
-----------------
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
-----------------
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
-----------------
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.
-----------------
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
-----------------
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
-----------------
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
-----------------
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
-----------------
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
