For the answer to the question above, I'll provide my solutions to my answers for the problem below.
(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)
(−2x3)(y2)+4x2y3+−3xy4+−1(6x4y)+−1(−5x2y3)+−1(−y5)
(−2x3)(y2)+4x2y3+−3xy4+−6x4y+5x2y3+y5
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
(−6x4y)+(−2x3y2)+(4x2y3+5x2y3)+(−3xy4)+(y5)
−6x4y+−2x3y2+9x2y3+−3xy4+y5
So the answer is,
= <span><span><span><span><span>−<span><span>6x4</span>y</span></span>−<span><span>2x3</span>y2</span></span>+<span><span>9x2</span>y3</span></span>−<span>3xy4</span></span>+y5</span>
I hope this helps
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: I got C.
Step-by-step explanation: I squared both 3.9 and 9.7 which gave me 109.3. Then, I square rooted it and got 10.45 which rounds to 10.5. Hopefully, this is right.