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:
128
Step-by-step explanation:
all angles of a triangle will always equal 180
so, we know 2 angles
180-40
140-1128
Y=-3x-2
subsitute -3x-2 for y
-7x+3(-3x-2)=10
distribute
a(b+c)=ab+ac
3(-3x-2)=-9x-6
-7x-9x-6=10
add like terms
-16x-6=10
add 6
-16x=16
multiply -1
16x=-16
divide 16
x=-1
subsitute
y=-3x-2
y=-3(-1)-2
y=3-2
y=1
x=-1
y=1
(x,y)
(-1,1)
Answer:
Step-by-step explanation:
We use binomial expansion for 
This can be rewritten as
![[x(1+\dfrac{h}{x})]^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5Bx%281%2B%5Cdfrac%7Bh%7D%7Bx%7D%29%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
From the expansion
Setting 
and 
,
Multiplying by 
,
The limit of this as 
is
(since all the other terms involve 
and vanish to 0.)
Answer:
12x-6
Step-by-step explanation:
