Alan

The following graph should help:

A and B must be integers in the range 0 to 8  (I'll assume here that A and B are both non-zero)

AB_9  is   9A+B

B0_9  is   9*B+0

So    9A+B + A = 9B + 0   or  10A = 8B

The only combination of integers in the range 1 to 8 that satisfy the above equality is A = 4 and B = 5, so A*B = 20 (in base 10).

How about the following:

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You have the following inequalities:

\(2x+1.5y\ge500\\0.25x+0.2y\le100\\x\ge150\)

Plug in the various pairs of x and y (eg x = 160, y = 110;  x = 232.5, y = 200; ...etc.) and see which pairs satisfy these inequalities.

(Strangely, it looks as though it's possible to sell a half cup of lemonade and half a lemon bar!  Unless, of course, they are clues to impossible outcomes!)

asin is the same as sin^-1

acos is the same as cos^-1

atan is the same as tan^-1

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The \(\Delta \)symbol means take the difference between successive values of ...

So \(\Delta f(x) \)means take the difference between successive values of f(x): namely 3 here, and \(\Delta x\)means take the difference between successive values of x, or 1 here.

So \(\frac{\Delta f(x)}{\Delta x}=\frac{3}{1}\rightarrow3\)

Like this:  \(\frac{(4x^3y^2)^3}{10}\rightarrow\frac{4^3(x^3)^3(y^2)^3}{10}\rightarrow\frac{64x^9y^6}{10}\rightarrow\frac{32x^9y^6}{5}\)

This doesn't involve proving trig identities, but using them to find the value(s) of \(\theta\)  and x that make the equations true.

By definition: \(\sec\theta=\frac{1}{\cos\theta}\)

Using this in a and rearranging:   \(\frac{1}{\cos^2\theta}=2\tan\theta\)

Multiply both sides by \(\cos^2\theta\) and using the fact that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) we get, on simplifying: \(1=2\sin\theta\cos\theta\)

The RHS is just \(\sin{2\theta}\)  so we have \(1=\sin{2\theta}\)

This results in \(2\theta=\frac{\pi}{2}\text{ or }\theta=\frac{\pi}{4}\) 

For b rewrite x as \(2\theta\) and use \(\cos{2\theta}=\cos^2\theta-\sin^2\theta\text{ and }\sin{}2\theta=2\sin\theta\cos\theta\text{ together with }\cos^2\theta+\sin^2\theta=1\)

You should eventually find there are two values of x between 0 and 2pi that satisfy the equation.

Remember that \(\cos^2x+\sin^2x=1\)

Rearrange this in the form:  \(\sin^2x=1-\cos^2x\)

Substitute this into the left hand side (LHS) of your identity:

LHS:  \(1-\cos^2x-cos^2x\)

Simplify

LHS: \(1-2\cos^2x\)

So LHS = RHS

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I agree:

w can only be 6

x can be 22  or  34

x = 22:  y can be 14 or  58        x = 34: y can be 2 or 70

y = 14: z can be 62 or 90      y = 58: z can be 18 or 134

y = 2:  z can be 74 or 78       y = 70: z can be  6 or 146

Four greatest values of z are 78, 90, 134, 146

78+90+134+146 = 448