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.
Substitute x for -1 in <span> y=3x-1
</span>y = <span><span>(3) <span>(-1) -</span></span><span>1
</span></span>Solve
y = (3) (-1) -1
y = -3 - 1
y = -4
B.) -4 is your answer
Answer:
The plane PRS passes through the points P, R and S. So it contains the line RS. Also the plane QRS passes through the points Q, R and S. So it contains the line RS as well. Since both the planes contain the line RS, the line RS must be the intersection of plane PRS and QRS
Step-by-step explanation:
Answer:
Step-by-step explanation:
If the equations are true, they can be solved simultaneously.
Consider 2x-10y=-1 as Eq1 and 5x+6y=4 as Eq2
Multiplying Eq1 with 3 and Eq2 with 5 we get,
6x-30y=-3 -- Eq3 and 25x+30y=20 -- Eq4
Adding Eq3 and Eq4,
31x=17 Therefore, x=17/31
Plugin x=17/31 in Eq3,
6(17/31)-30y=-3 --- y=13/62
Plug In values of x and y in 7x-4y we get,
7(17/31)-4(13/62) = 3
So the value of 7x-4y = 3.
