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
Answer:
P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
Step-by-step explanation:
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triangle.
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
hope this helped
Answer:
Step-by-step explanation:
we just multiply numerator by numerator and denominator by denominator
3/4x22/7
66/28x2/1
132/28
4 20/28
4 5/7
Hopes this helps,please mark brainliest
Do you remember when you were learning about
Least Common Multiple (LCM) ? And you were
wondering what on Earth you would ever need it for ?
THIS IS IT !
The amount of time after 8 PM when the lights will flash
together again is the LCM of 20 seconds and 30 seconds.
Do you remember how to find it ?
... When 20 is prime-factored, you have 2 · 2 · 5 .
... When 30 is prime-factored, you have 2 · 3 · 5 .
... So the Least Common Multiple of 20 and 30 is 2 · 2 · 3 · 5 = 60 .
The lighthouses will flash together again in 60 seconds after 8 PM.
That'll be 8:01 PM.
