The cost of using 19 HCF of water is $32.49
Given in the question:
The monthly cost (in dollars) of the water use (in dollars) is a linear function of the amount of water used (in hundreds of cubic feet, HCF)
The cost for using 17 HCF of water is using $32.13
and, the cost of using 35 HCF is $61.83.
To find the cost of using 19 HCF of water.
Now, According to the question:
The cost for using 17 HCF of water is $32.13
and, the cost of using 35 HCF is $61.83.
To find the slope:
(17, 32.13) and (35, 61.83)
Slope = (61.83 - 32.13)/ (35 - 17) = 1.65
We know that:
Formula of slope :
y = mx + b
32.13 = 1.65 x 17 + b
b = 1.14
The equation will be :
C(x) = 1.65x + 1.14
Now, To find the cost of using 19 HCF of water.
C(19) = 1.65 × 19 + 1.14
C(19) = $32.49
Hence, the cost of using 19 HCF of water is $32.49.
Learn more about Slopes at:
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Answer:
y = -x - 4
Step-by-step explanation:
the equation of a linear graph is y = mx + b
you know the m = -1
so now the equation is y = -x + b
from here, what you could do to find b is...
plug (-1, -3) into the equation
1) x = -1, y = -3
2) -3 = -(-1) + b
3) -3 = 1 + b
4) -4 = b
or you can look at the y-intercept which is where the line intersects with the y-axis
it seems to be (0, -4)
so "b" would be equal to -4
this makes the final equation y = -x + (-4) or y = -x - 4
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
