MaxWong

Using binomial theorem, \(\newcommand{\dsum}{\displaystyle\sum} (2y - 7)^5 = \dsum_{k = 0}^5 \binom{5}k (2y)^k (-7)^{5 -k}\)

To find the term in y^4, the power needs to be 4. So let k = 4:

\(\text{term in }y^4 = \displaystyle\binom{5}4 (2y)^4 (-7)^1 = -560y^4\)

Coefficient of y^4 = -560.

The interior angles of a triangle sums up to 180 degrees, so \(C = 180^\circ - 42^\circ - 75^\circ = 63^\circ\).

Then you just use a calculator to calculate \(\sin C = \sin 63^\circ\).

Hints:

1) How can you relate the rate of change in area to the rate of change in radius?

2) What is the radius when the area is 10 mi^2? This is the value of the radius you need to substitute when you substitute area = 10 mi^2.

When \(f(x) \geq 4\), \((x^2 - 4 \geq 4 \text{ and }x \geq 4) \text{ or } (x^2 + 4 \geq 4 \text{ and }x < 4)\), which is \(x < 4\text{ or }x \geq 4\), which means f(x) is always greater than or equal to 4.

Then f(f(x)) = (f(x))^2 - 4 by definition of f.

Case 1: x >= 4

\((x^2 - 4)^2 - 4 = 5\\ x^4 - 8x^2 + 16 - 4 - 5 = 0 \\ x^4 - 8x^2 + 7 = 0\\ (x^2 - 7)(x^2 - 1) = 0\\ x = \pm \sqrt 7 \text{ (rej.) or }x = \pm 1 \text{ (rej.)}\)

The roots are rejected since none of them are greater than 4.

Case 2: x < 4

\((x^2 + 4)^2 - 4 = 5\\ x^4 + 8x^2 + 16 - 4 - 5 = 0\\ (x^2 + 1)(x^2 + 7) = 0\\ \text{No real solution}\)

Therefore, there are no value of x that would make f(f(x)) = 5.

Sketch of solution:

Suppose the rate at which water is being pumped into the tank is k cm^3 / min. Then the net rate of increase of volume is (k - 6700) cm^3 / min.

You can do what I did in the previous problem (considering the largest cross-section) to find the radius of the water surface when the height is h cm. Draw a diagram to help you visualize the problem. When you have that, you can express the volume V entirely in terms of h. Differentiate both sides with respect to time and then you can relate the rate of change in volume to the rate of change in water level. That rate of change in volume when h = 2 is exactly (k - 6700) cm^3 / min.

Please tell me if you encounter problems solving it according to my sketch.

This problem is very much like this one here. If you understood how to do that one, you know how to do this one.

https://web2.0calc.com/questions/domain_89573

Anything on the denominator must not be equal to 0 and anything inside a square root must be nonnegative.

Therefore \(x^2 - 7x + 6 > 0\). If you solve the inequality, you get the domain of the function.

Suppose r is the base radius of the cylinder inscribing the cone. Let h be the height.

Using similar triangles on the largest cross-section, 

\(\dfrac{h}{8 - r} = \dfrac{6.5}8\\ h = \dfrac{13}2 - \dfrac{13}{16}r\)

Then the volume is \(\pi r^2 h = \dfrac{13}{16} \pi r^2 (8 - r)\).

Solving \(\dfrac{d}{dr} \dfrac{13}{16} \pi r^2 (8 - r) = 0\) gives r = 16/3 or r = 0 (rej.)

Note that 

\(\dfrac{d^2}{dr^2}\Big|_{r = \frac{16}3} \dfrac{13}{16} \pi r^2 (8 - r) = -13\pi < 0\). Then maximum volume is attained when r = 16/3.

Now, you can calculate the height and get your answer.

Constant term is \(\displaystyle \binom{9}3 (2z)^3 \left(-\dfrac1{\sqrt z}\right)^6\). You can simplify this to get the constant term 672.

By the way, I would love to see a solution without using calculus, since calculus is invented after Tartaglia.