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Artist 52 [7]
3 years ago
9

Find the measure of x, then find the measure of each angle​

Mathematics
1 answer:
stepladder [879]3 years ago
7 0

First, you have to imagine that you cut the upper triangle. If you do that, then this is how the inner angles are added to 180°:

180° = 90° + (x/2) + (7x - 90°)

180° = x/2 + 7x

180° = x *(1/2 + 7)

180° = x*(15/2)

2/15 * 180° = x

24 = x

Angle A:

A= x/2

A = 24/2

A = 12°

Angle B:

B = 7x

B = 7*24

B = 168°

The two bottom angles are both worth 90°.

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6 0
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2/5da80% como ahora la cuenta
andreev551 [17]
I don't speak Spanish but I think that this is what you meant:
80 * 2/5 = 32
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4 0
3 years ago
Need help with the blanks
Crazy boy [7]
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
3 0
3 years ago
Plz help me in here and tysm for ya help :3
lapo4ka [179]
It would cost somehow less than $1000 and you would know that because 347 students is a large number, but since it hold 44 students in each bus, it’ll take not to long to get to 347. So since each bus is $95, it would at least reach to 700+

(All is estimated)
8 0
3 years ago
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