Answer:
a) The estimates for the solutions of 
are 
and 
.
b) The estimates for the solutions of 
are 
and 
Step-by-step explanation:
From image we get a graphical representation of the second-order polynomial 
, where 
is related to the horizontal axis of the Cartesian plane, whereas 
is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:
a)
There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:
,
b)
There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:
,
<em>x equals 22</em>
<h2>
Explanation:</h2>
____________________________________________
The Segment Postulate states the following:
<em>Given two end points A and B, a third point C lines on the segment AB if and only if the distances between the points satisfy the equation:</em>
____________________________________________
From the figure:
Our goal is to find x:
<h2>Learn more:</h2>
Dilation: brainly.com/question/2501119
#LearnWithBrainly
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
-----------------
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
-----------------
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
The common point between y = 2x + 5 and y = (1/2)x + 6 will have the same values for x and y.
Therefore, set the two y expressions equal to obtain
2x + 5 = (1/2)x + 6
Subtract (1/2)x from each side.
(3/2)x + 5 = 6
Subtract 5 from each side.
(3/2)x = 1
Multiply each side by 2/3.
x = 2/3.
From the first equation, obtain
y = 2*(2/3) + 5 = 19/3.
The common point is (2/3, 19/3). It is not equal to (3, 1/2).
Answer: False
