Because of the vertical angles theorem, BAC becomes 50°. And angle B is 180-(50+30)=100.
If all 3 triangles are similar, the angles of triangle EFG are 30°, 50°, 100°
Answer:
Z = 8 + 2x2 + 2y2
Convert to polar coordinates
Z = 8 + 2r2
Now theta will go from 0 to pi/2 because it's in the first quadrant.
R will go from 0 to the radius of the circle formed at the intersection of the plane and the paraboloid.
14 = 8 + 2r2
r = sqrt(3)
So r goes from 0 to sqrt(3).
You integrate 14-z where 0<r<sqrt(3) and 0<theta<pi/2.
It is 14-z and not z because just z would give the volume under the paraboloid.
Step-by-step explanation: please go answer my recent question
The lines y=x+2 and y=x are parallel, because they have the same slope.
Answer/Step-by-step explanation (ac > b² or b² < ac.
)
A/c to question, we have to show:-
b² >ac in A.P ........ (1)
b² = ac in G.P .....(2)
b² < ac in H.P. ..... (3)
b = a+c/2 (A.P)
b = √ac ( G.P)
b = 2ac/a+c (H.P)
In A.P :
b² > ac = b² - ac
= (a+c/2)² - ac
= (a²+2ac+c²/4) - ac = a² + 2ac + c² - 4ac / 4
= a² - 2ac + c² / 4 = ( a - c ) ² / 4 > 0 Hence, b²>ac
In G.P:-
b = √ac
Hence, b² = ac
In H.P :- b² < ac = ac > b² = ac - b² = ac - ( 2ac / a+c)
= ac(a+c) - 2ac / a+c
= a²c + ac² - 2ac / a+c
= ac(2ac - 2) / a+c > 0
Hence, ac > b² or b² < ac.
We have the following equation:

If we graph this equation we realize that in fact this is an ellipse with
major axis matching the y-axis. So we can recognize these characteristics:
1. Center of the ellipse: The midpoint C<span> of the line segment joining the foci is called the </span>center<span> of the ellipse. So in this exercise this point is as follows:
</span>
2. Length of major axis:
The line through the foci is called the major axis<span>, so in the figure if you go from -5, at the y-coordinate, and walk through this major axis to the coordinate 1, the distance you run is the length of the major axis, that is:</span>
3. Length of minor axis:
The line perpendicular to the foci through the center is called the minor axis. So in the figure if you go from -2, at the x-coordinate, and walk through this minor axis to the coordinate 2, the distance you run is the length of the minor axis, that is:
4. Foci:Let's find c as follows:

Then the foci are:
