Answer:
C is a constant ( a value that doesn't change)
Step-by-step explanation:
for example (X^2), here x is a variable with an exponent. (8x), here 8 is a coefficient. making c a constant.
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: Number one is C and number two is B
Let n be a number of balls: n = 14
In one pan you have 8 balls: 8n
In the other pan you have the rest of balls (14 - 8 = 6) along with a <span>weight of 20 grams = 6n + 20
So the pans are balanced:
8n = 6n + 20
Now, just solve the equation:
8n - 6n = 20
2n = 20
n = 20/2
n = 10
So, you know that a ball weights 10 grams.</span>
Answer:
DE ≈ 14.91
Step-by-step explanation:
Make use of the relationships between sides and angles in a right triangle. These are summarized by the mnemonic SOH CAH TOA:
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
Tan = Opposite/Adjacent
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The side DE is opposite the angle 19°, so the sine or tangent relation will be involved. The sine relation requires you know hypotenuse EF. The tangent relation requires you know adjacent side DF.
The only common side between triangles CDF and DEF is side DF. That side is opposite the given 61° angle. The given side length (CF = 24) is adjacent to the 61° angle.
This means you have enough information to use these relations:
tan(61°) = DF/CF = DF/24
DF = 24·tan(61°)
and
tan(19°) = DE/DF
DE = DF·tan(19°) = (24·tan(61°))·tan(19°) . . . . . use DF from above
DE ≈ 24(1.804048)(0.344328) ≈ 14.908
The length of DE is about 14.91.
