For a better understanding of the explanation provided here, please find the diagrams in the two files that have been attached.
As can be clearly seen from the diagram, PR and QS are the diameters of the circle and they intersect each other at the centre, O of the circle.
The angle these diameters make at the point of intersection O is 90 degrees.
If Margot joins the points P,Q,R and S which are on the circumference of the circle, then we will get a square with sides PQ, QR, RS and SP as shown in the second diagram.
Thus, the correct option is Option A.
At the beginning, there are 15 pigs without ribbons. one of the pigs receives a ribbon. how many are left ribbonless?
<span>14. when another pig is given a ribbon, there are 13 left. after all the ribbons have been given out, 3 of the 15 pigs have ribbons. </span>
<span>if any one of the 15 at the beginning is given a ribbon, there are 15 possible combos. for the second ribbon, there are 14 combos- but each of those combos has to go through the first 15. for the third ribbon, there are 13 combos, going through 14 and 15. </span>
<span>simply put, 15 x 14 x 13.</span>
Answers:
21. Three noncollinear points determine 3 lines and 1 plane
22. If two quadrilaterals are similar, then they are squares
23. PR = 2.6
24. Midpoint is (3,-1)
25. Angle BXD = 108 degrees
26. The complement is 18 degrees
27. A) 30
28. Corresponding angles
29. False; Change "congruent" to "supplementary"
30. Neither
31. Equation is y = (-2/5)x + 9/5
32. Obtuse scalene triangle
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Work Shown
Problem 21)
Three noncollinear points determine 3*2 = 6 pairings but half of those pairings are repeats, so we have 6/2 = 3 unique groups forming 3 lines (think of a triangle and its sides)
The three noncollinear points form a single plane. This is simply an axiom.
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Problem 22)
Original Conditional is in the form If P, then Q
The converse is the flip of that. So we go to If Q, then P.
So we have
Original Conditional: "If two quadrilaterals are squares, then they are similar"
Converse: "If two quadrilaterals are similar, then they are squares"
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Problem 23)
P is between Q and R. By the segment addition postulate, we know
QP+PR = QR
We're given PQ or QP to be 10.2 and we know that QR = 12.8, so this means,
QP+PR = QR
10.2+PR = 12.8
10.2+PR-10.2 = 12.8-10.2
PR = 2.6
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Problem 24)
Add up the x coordinates and divide by 2: (x1+x2)/2 = (8+(-2))/2 = 6/2 = 3
Add up the y coordinates and divide by 2: (y1+y2)/2 = (-6+4)/2 = -2/2 = -1
Therefore the midpoint is (3,-1)
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Problem 25)
Angle DXE = 36 (given)
Angle CXD = angle DXE (definition of bisection)
Angle CXD = 36
Angle CXE = (angle CXD)+(angle DXE)
Angle CXE = 36+36
Angle CXE = 72
Angle BXE = 2*(angle CXE) ... since XC bisects angle BXE
Angle BXE = 2*72
Angle BXE = 144
Angle BXD = (angle BXE) - (angle DXE)
Angle BXD = 144 - 36
Angle BXD = 108
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Problem 26)
From problem 25, we found that Angle CXE = 72. Since XC cuts angle BXE in half, and the other angle is BXC, this means
Angle BXE = angle CXE = 72 degrees
Now subtract that from 90
90 - (angle BXE) = 90 - 72 = 18
The complement is 18 degrees
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Problem 27)
A+B+C = 180
x+x+120 = 180
2x+120 = 180
2x+120-120 = 180-120
2x = 60
2x/2 = 60/2
x = 30
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Problem 28)
Angle 5 and angle 7 are corresponding angles. They are located on the same side of the transversal line. They both correspond to the same side of their respective parallel line counterparts. Both are on the right side of the parallel line they are attached to.
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Problem 29)
The statement in its current form is False. One way to fix it is to change the first underlined term from "congruent" to "supplementary". Angle 3 and angle 2 are same side interior angles which add up to 180 degrees.
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Problem 30)
Slope of AB = (y2-y1)/(x2-x1)
Slope of AB = (-2-6)/(-2-10)
Slope of AB = -8/(-12)
Slope of AB = 2/3
Slope of CD = (y2-y1)/(x2-x1)
Slope of CD = (2-6)/(6-(-6))
Slope of CD = -4/12
Slope of CD = -1/3
Multiply the slopes:
(Slope of AB)*(Slope of BC) = (2/3)*(-1/3) = -2/9
The result is NOT equal to -1, so the lines are NOT perpendicular
The two slopes are NOT equal, so the lines are NOT parallel
So the answer is "neither"
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Problem 31)
Anything parallel to 2x+5y = 12 is of the form 2x+5y = C where C is some fixed number
Plug in the given point (x,y) = (2,1) to find C
2x+5y = C
2*2+5*1 = C
4+5 = C
9 = C
C = 9
So we go from 2x+5y = C to 2x+5y = 9. Now solve for y
2x+5y = 9
2x+5y-2x = 9-2x
5y = -2x+9
5y/5 = (-2x+9)/5
y = (-2/5)x + 9/5
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Problem 32)
A+B+C = 180
16+B+64 = 180
B+80 = 180
B+80-80 = 180-80
B = 100
The angle B is 100 degrees, which is larger than 90 degrees. We have an obtuse triangle because of this fact.
All three angles (16, 64, 80) are different, so the side lengths are different. The three different side lengths means we have a scalene triangle.
If the hypotenuse is equal to 11, then we can plug it into the Pythagorean theorem. So we get a^2+x^2=121. So we take the square root of both sides, sqrt(a^2+x^2)=11. So we take a^2 and x^2 out of the square root. We get a+x=11. So now we have two equations, a+x=11 and a^2+x^2=121. I’ll leave the rest of solving equations to you.
Hint: a=11-x
You can plug that into a^2+x^2=121
Answer:
scale factor = 2
Step-by-step explanation:
The scale factor is the ratio of corresponding sides, image to original
Using the bases of the 2 triangles , then the scale factor is
S → L = 
= 2
