The points where the 2 graphs intersect is where x = 0 and x = 2.
- 2 2
x = INT x dA / INT dA
0 0
INT dA = INT -x^2 + 4x + 3 - (x^2 + 3 ) dx = INT -2x^2 + 4x
= -2 x^3/3 + 2x^2
= 2.667 between 0 and 2
xdA = -2x^3 + 4x^2 INT xdA = -x^4/2 + 4x^3/3 = 2.667
centroid = 2.667 / 2.667 = 1 (x = 1)
Answer:
x=16 and y=9
Step-by-step explanation:
2y+30=3y+21[Diagonals of parallelogram are equal]
30-21=3y-2y
y=9
3y=2x-5[Diagonals of parallelogram are equal]
3×9=2x-5
27=2x-5
2x=27+5
x=32/2=16
Given:
M=(x1, y1)=(-2,-1),
N=(x2, y2)=(3,1),
M'=(x3, y3)= (0,2),
N'=(x4, y4)=(5, 4).
We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.
For a parallelogram, opposite sides are equal
If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.
To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,
Slope of MN= Slope of M'N'.
Slope of MM'=NN'.
Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'
Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.
Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.
Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.
Remark
There's a lot you don't know here. Are DE and GF parallel? Is B a right angle? You can't assume that it is. The safest way to proceed is to give x in terms of 58 and B. You might get an answer that gives you something like 32 but I don't think you can say that unless you are told somewhere that ABC is a right angle triangle with the right angle at B.
So what to do.
<BAC = 58o That's because <BAC = <IAK They vertically opposite.
<ABC + <BAC + <ACB = 180o All triangles have 180o
<ACB = 180 - 58 - <ABC Solve for an unknown angle of a triangle.
<ACB = 122 - <ABC
x = <ACB Vertically opposite angles.
x = 122 - <ABC Answer It's 32 if ABC is a right angle.
