Answer:
y = 4x - 12
Step-by-step explanation:
Please let me know if you want me to add an explanation as to why this is the answer/how I got this answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)
F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)
d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.
If in 2 hours he read 200 pgs then in 1 hour will be 50 pgs
Part A.
The trip starts at 8am which corresponds to 0 hrs, point (0hr, 0mi)
2hrs later it's 10am. .point (2hr, 140mi)
The average speed is the slope between 0 and 2 hrs. Remember the slope formula m = Δy/Δx
m = (140 - 0) / (2 - 0)
m = 70 mph
Part B. Average speed from 11am - 2pm
11am is point (3hr, 140mi)
2pm is point (6hr, 300mi)
As you can see from the graph, the speed or slope changes at 1pm (5,260). You Can just use the start and end points.
m = (300-140) / (6-3)
m = 160/3
53.3 mph
* It comes out the same solution as if you average the two different slopes. 2hrs at 60mph + 1 hr at 40mph = (120 + 40)/3 = 160/3
Part C. Total average speed = total distance / total time driving
He went 70 mph for 2 hrs
stopped for an hour (slope is zero, no speed)
60 mph for 2hrs
40mph for 1 hr
300mi /5hr = 60mph
Part D. No Question....
To find the probability of landing on a triangle, you will want find the combined areas of the triangles and the total area of the square target.
Divide the area of the combined areas and the total area to find the probability of landing on a triangle.
A = 1/2bh
1/2 x 8 x 8
A = 32 square inches
32 x 4
128 square inches (areas of triangles)
A = bh
26 x 26
A = 676 square inches
128/676 = 0.189
There is an approximate probability of 0.19 of hitting a triangle.
