Find y': ○ 4kx (2kx + 1)7 ○ 4k (2kx + 1)7 ○ 8k(kx² + 1)7 ○ 8kx(kx² + 1)7 y = 1/(kx² + 1)8 ; k constant

Q: d^2B(u)/du^2 ,but where B(u)= 1/(1+u)^7

A: Given B(u)=1(1+u)7To find d2[Bu]du2:

Q: If an object is thrown directly from t seconds s (t) = -16t ^ 2 + 48t + 256 feet after t seconds.…

A: For maxima, [double derivative of  s(t)]<0.

Q: (*+)(LX +ン 2" (2n +1)

A:

Q: Find the areas of the shaded regions in

A:

Q: Evaluate I = = / x*(x² + 4)³/ª dz

A:

Q: y=3+1/x^2, compressed horizontally by a factor of 2

A: Given query is to apply the Indicated transformation.

Q: p(t)=-1718^2+82000+10000

A: Given:p(t)=-17182+82000+10000

Q: 17x^(2)+7y^(2)=119. Find the length of a, b, c.

A:

Q: P(X)=x^(3)-4x^(2)-8,P(2)

A: given that  p(x) = x3-4x2-8 then  P(2)=?

Q: y = (x' - 2x2 +3) (7x² – 4x)

A:

Q: 4sin^(2)(x)-3=0

A:

Q: (10x-19) (3r + 2) (5x-3)

A: We will use the exterior angle property of a triangle that states that an exterior angle is equal to…

Q: 4x/(5x -- 25) -- 10/(x -- 5) = x + 4/5       x=?

A: The given equation is

Q: (x^(2)-4y)/(2) x=4 y=-3

A:

Q: p(x) = 1/(3x ^ 3) + 2/(3x ^ 2) + 3y Find p(- 2) and p(3) for function

A: Given: px=13x3+23x2+3y Now for x=-2, we get,…

Q: Find h(-2)         h(t)=t^2-4t+2

A: The given function, h is,

Q: f(x)= 3(x+5)^2-1; (-4,2)  what's the axis of symmetry

A: Solution: For a quadratic function in a standard form ax2+bx+c The axis of symmetry is a vertical…

Q: D^(3+1) (D-2)^(7+1) y = 48e^2x

A: Put D=m then solve it.

Q: ({)*** = 2x+2 27 %3D 125

A: topic - exponents and linear equation

Q: Divide f (x) = 2x ^ 4 + 3x ^ 2 + 10 by g (x) = 3x ^ 2 + 5x + 1 in Z11.

A: Solution:         To Divide f(x)=2x4+3x2+10  by g(x)=3x2+5x+1…

Q: y=(x+x*}

A:

Q: x^2=Axy+By

A: Simple value put

Q: Find domain and range of the * function y = Vx2 – 4x

A: To find the domain and range of  y=√(x²-4x)

Q: Differentiate. h(w) (w5+ 8w^) (w-4-w-6)

A:

Q: =a(x-1)^2+2 passes through the points (10,-160) A=

A: NOTE: Refresh your page if you can't see any equations. . the point (10,-160) is on the graph so…