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The inverse function will have its domain restricted to values of x \ge 1. Why?

note … \sqrt{\sin^2{x}} = |\sin{x}|

cost of the lead lining is proportional to the surface area of the open container if h = 3, then A = 144 if h = 1, then the area of the bottom alone is 144 … adding the area of the sides would make A > 144 I’d say you are correct.

start by multiplying every term by 4 ... 8 \cdot 2^{2x} + 32 \cdot 2^x = 4 + 2^x

u=x^3+x^2+x \implies du =3x^2+2x+1 note … 6x^2+4x+2 = 2 \, du

Shown is graph of the basic secant function in black with domain restricted to \left[0,\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\pi\right] so that it passes the horizontal line test for the existence of an inverse function. Reflection over the line y = x is the inverse secant function in...

it hasn't changed in the 20+ years I've taught the course ... \left(\dfrac{\text{top}}{\text{bottom}}\right)' = \dfrac{\text{bottom times derivative of the top} - \text{top times derivative of the bottom}}{(\text{bottom})^2}

because that's not the quotient rule ... \dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}

\dfrac{d}{dx} \bigg[\dfrac{\sin{x}}{1+\cos{x}}\bigg] = \dfrac{(1+\cos{x})\cos{x} - \sin{x}(-\sin{x})}{(1+\cos{x})^2} try again ...

sketch a free body diagram of all forces acting on the crate. note the crate is in a state of equilibrium.

For part (a), refer to the 3D diagram. Note the right triangle with hypotenuse R, vertical leg y/2, and horizontal leg \dfrac{\sqrt{2}}{2} \cdot x

geometric mean ...

\displaystyle \int_0^{\pi/2} \dfrac{\cos{x}}{\sin{x}(1+\sin{x})} \, dx use the substitution u = 1+\sin{x}, get an integrand in terms of u, and reset the limits of integration. From there, I would recommend using a partial fraction decomposition.

On the muscle of my arm is a red & blue tattoo … sez “Fort Worth I Love You”. It’s “home”, but not the only place I’ve resided after 20+ years in the USN.

no dodge that I can see y' = 2x+4 y'(3) = 10

city A is East of city B by two time zones ... flying East to West one gains two local hours at destination; flying West to East one loses two local hours at destination

In the horizontal direction, force components to the right = force components to the left R\cos(35) = 11 + Q\cos(45) In the vertical direction, force components up = force components down R\sin(35) + Q\sin(45) = 14 solve the system of equations for R and Q in kN.