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First, manoeuvre variables so that we can integrate the equation.
1/(x-6)^(1/2) dx = -2 dt
Integrate the equation and add the constant.
2(x-6)^(1/2) = -2t +c
Solve for t.
t = -(x-...

2(x-2)^2 +13

first we know that the length given has a 4:1 ratio with the remaining length of the radius so 5/4 + 5 = 25/4 = radius, the formula for the area of a circle is A=pi*(r^2) so the area is 625/16*pi.

Using the substitution u = sec(z)=> du = sec(z)tan(z) dz.So, the integral ∫ y dz = ∫ sec(z)tan(z)/sqrt(sec(z)) dz=> ∫ y dz = ∫ 1/sqrt(u) du = 2sqrt(u) + C = 2sqrt(sec(z)) + C.

First, you will need to differentiate the function with respect to x. Finding dy/dx.For polynomials, this is done by taking one away from the old power and multiplying the coefficient by the old power and...