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Mariana [72]
3 years ago
8

A student's work to solve an equation is shown.

Mathematics
2 answers:
Leto [7]3 years ago
8 0

Answer:

Incorrectly , one solution

Step-by-step explanation:

The final solution of student is 2 = 2 which is a true statement, i.e. equation has infinitely many solution,

Given equation,

\frac{1}{8}(40x+16)=9x-7(2x-1) -5

5x + 2 = 9x - 14x + 7 -5 ( by distributive property )

5x + 2 = -5x +2

5x + 5x = 2 - 2

10x = 0

\implies x = 0

Hence, the equation has a solution which is 0.

Therefore, the student solved the equation incorrectly because the original equation has one solution(s).

zvonat [6]3 years ago
6 0

Answer:

the student solved the equation incorrectly because the original equation has one solution

Step-by-step explanation:

1/8(40x + 16) = 9x − 7(2x − 1) − 5

<em>use the distributive property a(b + c) = ab + ac</em>

(1/8)(40x) + (1/8)(16) = 9x + (-7)(2x) + (-7)(-1)

5x +2 = 9x − 14x + 7 − 5       <em>combine like terms</em>

5x + 2 = (9x - 14x) + (7 - 5)

<em>here the student make mistake 9x - 14x = 5x</em>

5x+2=-5x+2    <em>add 5x to both sides and subtract 2 from both sides</em>

5x + 5x + 2 - 2 = -5x + 5x + 2 - 2

10x = 0          <em>divide both sides by 10</em>

10x/10 = 0/10

x = 0

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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
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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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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
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5x/5 = 270/5
x = 54
Use this to find L
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L = 2*54
L = 108
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2x+40 = 180
2x+40-40 = 180-40
2x = 140
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(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
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