Answer:
x = 20
Step-by-step explanation:
x/10 = 2 multiply both sides of the equation by ten ↓
10x x/10 = 10 x 2 reduce the numbers with the greatest common factor 10 ↓
x = 10 x 2 multiply the numbers ↓
x = 20
Answer:
7.) 7
10.) 0
Step-by-step explanation:
When it means "evaluate the function", it's in essence asking us to see what the function spits out when we feed it a certain input. Our inputs are our x values, which spit out a y value.
Evaluating the function when x = 1:
Let's look at where the function has an x value of 1. We see it near the bottom of the table and see the y value associated with the input is 7. So when the function is fed 1 as an input, it spits out 7.
Evaluating the function when f(x) = - 2:
This one is a weird because of the new notation. Just think of it as some value of f, which we don't know (so we represent it as an x-variable) must equal -2. So let's look at our table to find out where our output is -2. We find that when f(x) = -2 the input is 0. So the input which gives -2 is 0.
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:
The other end point is: s+ti = 3+9i
Step-by-step explanation:
Mid-Point(M) in the complex plane states that the midpoint of the line segment joining two complex numbers a+bi and s+ti is the average of the numbers at the endpoints.
It is given by: 
Given: The midpoint = -1 + i and the segment has an endpoint at -5 - 7i
Find the other endpoints.
Let a + bi = -5 -7i and let other endpoint s + ti (i represents imaginary )
Here, a = -5 and b = -7 to find s and t.
then;
[Apply Mid-point formula]
On comparing both sides
we get;
and 
To solve for s:
or
-2 = -5+s
Add 5 to both side we have;
-2+5 = -5+s+5
Simplify:
3 = s or
s =3
Now, to solve for t;
2 =-7+t
Add 7 to both sides we get;
2+7 = -7+t+7
Simplify:
9 = t
or
t =9
Therefore, the other end point (s+ti) is, 3+9i
