Alan

Click on the LaTex Formula button on the input ribbon and then insert your expression:

$$$729^{-\frac{2}{3}}$$$

y = 2x + 8

The slope is 2: for every added topping the value of y increases by 2.

The y intercept is the value of y when x = 0, so y_intercept = 8

The x intercept is the value of x when y = 0, so 0 = 2*x_intercept + 8.  rearrange to find x_intercept = -4

After 1 year the price is (1-0.08)*25000 = 0.92*25000.

After 2 years the price is (1-0.08)*0.92*25000 = 0.92²*25000

After 3 years the price is (1-0.08)*0.92²*25000 = 0.92³*25000

etc. until

After 6 years the price is (1-0.08)*0.92⁵*25000 = 0.92⁶*25000

$${{\mathtt{0.92}}}^{{\mathtt{6}}}{\mathtt{\,\times\,}}{\mathtt{25\,000}} = {\mathtt{15\,158.875\: \!033\: \!6}}$$

Hover the cursor over the log key and ln (the natural log) and log₂ should appear.

There are 12 possibilities for 1st place.  For each of these there are 11 possibilities left for 2nd place, so there are 12*11 possibilities for the first two places.  For each of these there are 10 possibilities left for 3rd place, so there are 12*11*10 possibilities for the first 3 places.

$${\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{1\,320}}$$

Try to find two numbers that multiply together to make 30 and also add together to make 11.

Look at the possible pairs of numbers that multiply together to make 30.  These are:

1  30

2  15

3  10

5  6

Which of these pairs when added makes 11?  Clearly, it is the last pair.  This means that your original equation can be written as (n+5)*(n+6) = 0, which means that there are two possible values for n, namely n = -5 and n = -6.

Chris wrote; 

This one would be difficult to solve without a solving program....although we could use some Calculus techniques to get an approximation, like Alan did. However, since you're asking a trig question, I might assume that you haven't taken Calculus, yet.

Using Wolfram, the solutions (in rads) are -2.13105 (-122.1°) and .585321 (33.54°)

We should note that Wolfram's results are also approximate.

An alternative, non-calculus based technique, would be a binary section technique.  Choose two values of opposite sign either side of a solution (use the graph to come up with initial choices).  Then halve the interval and see if f(x) is sufficiently close to zero. If it isn't but is positive, replace the previous positive value by this new one (similarly, if the new one is negative, replace the previous negative value by the new one) and keep repeating this process until f(x) is within some pre-specified tolerance of zero.

Plot a graph of the left-hand side:

It looks like there is a solution between x = 3 and x = 4.    Let f(x) = 5.4^x - 4.9^x - 105 and find the value of x that makes this zero.

You need to adopt a numerical method, like the ones noted here:  

The solution is (approx) x = 3.683447

Alan, did you do that whole answer in Texmaker and then copy it over?

I did (well, the LaTeX bit - not the graph!) and it didn't work perfectly (the first few lines aren't in the same position as in Texmaker)!

Let f(x)=x²+3sin(x)-2.  We want the value(s) of x that make f(x)=0.

First let's plot a graph of the function:

We can see there are two solutions to f(x) = 0.  Because of the sin(x) term we need to solve this numerically.  One way of doing this is to use the Newton-Raphson method:

$$\textbf{Newton-Raphson numerical method}.\\
$$f(x)=x^2+3sin(x)-2$$\\
$$f'(x)=2x+3cos(x)$$

Guess an initial value for $x$ call it $x_0$. Then iterate towards a solution as follows:
$$x_{n+1}=x_n-\frac{f(x_n}{f'(x_n)}$$ \\

Because there are two solutions here, we need to do this twice with a different guess each time. From the graph we can see that $x_0=-2$ and $x_0=1$ might be good guesses.\\
Because Newton-Raphson tends to converge quickly, we probably don't need to iterate many times. Here after about 4 iterations we converge on $$x=-2.1310748$$ and $$x=0.5853206$$ from our two guesses, which looks consistent with the graph above!$$