Alan

You missed a minus sign on the 2x gobromash.

.

Hmm!

.

I started by breaking 6x into 6(x^1/2)² and thinking of x^1/2 as k, so that h is then 6k², which leaves g as sin²h.

It is not a unique breakdown! For example we could also have k = (6x)^1/3, h = x^3, g = sin²x

or an infinite number of other possibilities!

.

One possibility is as follows:

\(k(x)=x^{1/2}\\h(x)=6x^2\\g(x)=\sin^2 x\)

Note that \(h(k(x))=6(x^{1/2})^2\rightarrow6x\)

.

.

Remember that sec(t) = 1/cos(t) so multiply numerator and denominator by cos(t), and note that tan(t) = sin(t)/cos(t) and that 1 - cos(t)^2 = sin(t)^2:

\(\frac{\tan t}{\sec t-\cos t}\rightarrow\frac{\tan t\times \cos t}{1-\cos^2t}\rightarrow \frac{\sin t}{\sin^2 t}\rightarrow \frac{1}{\sin t}\rightarrow \csc t\)

.

52 hours + (3 hours per week* w weeks) = (52 + 3w) hours

This must be at least 90 hours, so   \(52+3w\ge90\)

.

Typo in the last line Chris.  15 = 75/5 so:  PB = 75/5 - 27/5 = 48/5  

.

You scale u by -4 (i.e. you multiply each component by -4) and then add the result to v (i.e. just add corresponding components).

-4<-7, 1> + <3, 17>  →  <28, -4> + <3, 17> → <31, 13>

. 

Assume height above ground where arrow hits rock is s meters.  

Rock:   18.35 - s = (1/2)*9.8*t²      where t = time in seconds and accn of gravity = 9.8m/sec²

Arrow:  s = 47*t - (1/2)*9.8*t^2

Replace s in the first equation using the second equation:

18.35 - (47t - (1/2)9.8*t^2) = (1/2)*9.8*t²

18.35 - 47t + (1/2)*9.8*t^{2 =} (1/2)*9.8*t²

18.35 - 47t = 0

t = 18.35/47 ≈ 0.39 secs

.

The above uses the fact that for constant acceleration situations:

distance = initial velocity*time + (1/2)*acceleration*time²

??  !!!

.