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.
Answer:
y= -2x+15
Step-by-step explanation:
That question is impossible because we don’t know what we are trying to solve for it has to be x or y in this scenario it can’t be both we need one to find the other
Answer:
1. The product of thirty and eight more than fifty-two.
2. Thirty times the sum of eight and fifty-two
Step-by-step explanation:
1) 52+8 (eight more the fifty-two )
52+8 (The product of thirty and eight more than fifty-two.)
2) 52+8 (The sum of eight and fifty-two)
52+8 (Thirty times the sum of eight and fifty-two)
Hello,
7) A∪C={1,2,3,4,5,7,9}
8) A∩B={2,4}
C'= complement of C ={2,4,6}
9) A∪B∩C'={1,2,3,4,6,8}∩{2,4,6}={2,4,6}
10) A∪(B∩C')={1,2,3,4}∩{2,4,6}={1,2,3,4,6}
Are you blind?
