Answer:
We need the following three rigid motions:
i) Reflection around y-axis, ii) Translation three units in the -y direction, iii) Translation four units in the -x direction.
Step-by-step explanation:
We need to perform three operations on pentagon ABCDE to create pentagon A'B'C'D'E':
i) Reflection around y-axis:
(Eq. 1)
ii) Translation three units in the -y direction:
(Eq. 2)
iii) Translation four units in the -x direction:
(Eq. 3)
We proceed to proof the effectiveness of operations defined above by testing point D:
1) 
Given.
2) 
By (Eq. 1)
3) 
By (Eq. 2)
4) 
By (Eq. 3)/Result
Answer:
(-1, -1) Let me know if the explanation didn't make sense.
Step-by-step explanation:
If we graph the three points we can see what looks like a quadrilateral's upper right portion, so we need a point in the lower left. This means M is only connected to N here and P is only connected to N. So we want to find the slope of these two lines.
MN is easy since their y values are the same, the slope is 0.
NP we just use the slope formula so (y2-y1)/(x2-x1) = (-1-3)/(5-4) = -4.
So now we want a line from point M with a slope of -4 to intersect with a line from point P with a slope of 0. To find these lines weuse point slope form for those two points. The formula for point slope form is y - y1 = m(x-x1)
y-3 = -4(x+2) -> y = -4x-5
y+1 = 0(x-5) -> y = -1
So now we want these two to intersect. We just set them equal to each other.
-1 = -4x -5 -> -1 = x
So this gives us our x value. Now we can plug that into either function to find the y value. This is super easy of we use y = -1 because all y values in this are -1, so the point Q is (-1, -1)
the answer is $34. 36 if I understood your question correctly.
The two parabolas intersect for
and so the base of each solid is the set
The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, 
. But since -2 ≤ x ≤ 2, this reduces to 
.
a. Square cross sections will contribute a volume of
where ∆x is the thickness of the section. Then the volume would be
where we take advantage of symmetry in the first line.
b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of
We end up with the same integral as before except for the leading constant:
Using the result of part (a), the volume is
c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is
and using the result of part (a) again, the volume is
Answer:
64
Step-by-step explanation:
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