They can be different because Daryl is doing it online so he will not have to swing arcs on the computer screen he will just use circles. Hope that this helps!
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(11,1) because if you plug in x and y this answer is the only one that gives you 18 in the end
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:
$38
Step-by-step explanation:
Since the last two numbers aren't over 50, we would round down. This leaves you with $38.
8z +3 - 2z < 51
6z + 3 < 51
-3. -3
6z < 48
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6. 6
z<8
