Alan

Electromagnetic waves all travel at the speed of light in a vacuum, namely 299792458*m/sec, whatever their frequency.

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The y-intercept occurs when x = 0.

For the first pair you are told that when x = 0, y = 3, so the y-intercept is 3.

For the others (assuming a straight line is involved) find the slope from (y2-y1)/(x2-x1) and

equate this to (y2 - yintercept)/(x2-0)

So, for the second problem:  (-1 - 4)/(3 - -2) = (-1 - yintercept)/(3 - 0)

or:  -5/5 = (-1 - yintercept)/3

Simplify and multiply both sides by 3

-3 = -1 - yintercept

Add yintercept +3 to both sides to get:    yintercept = 2

I'll leave you to use the same technique on the third problem.

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Father's age = f, son's age = s

f - 4 = 7(s - 4)    Four years ago   (1)

f + 4 = 3(s + 4) Four years hence   (2)

From (1) we have  f - 4 = 7s - 28  or  f = 7s - 24  (3)

From (2) we have f + 4 = 3s + 12 or f = 3s + 8   (4)

From (3) and (4):   7s - 24 = 3s + 8  or  4s = 32 so s = 8

Put this back into (4) to get f = 3*8 + 8  or  f = 32

Father's age now is 32

Son's age now is 8

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Sigma notation explanation:

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It might help to think of a simpler situation first.  Consider four people sitting two each side of a two-sided table(!).

The distinct possibilities, excluding rotations, are easily visualised:

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This question might be related to the following common coin weighing problem.

You have 10 coins that look identical, but one is counterfeit.  Using a balance (i.e. where you put coins in pans at both ends of a "see-saw", but can't tell the actual weight) what is the minimum number of weighings required to identify the counterfeit coin (and can you tell if it is heavier or lighter than a genuine coin?)?

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In this case Melody, it might be better to let y = √x  (a positive number) when you get to 0.08x - 1 = √x  so that you have  0.08y² - 1 = y.  Then when you solve for y, if you get a negative number you can discard it immediately:

$${\mathtt{0.08}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{y}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,-\,}}{\mathtt{25}}\right)}{{\mathtt{4}}}}\\
{\mathtt{y}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.930\: \!703\: \!308\: \!172\: \!535\: \!8}}\\
{\mathtt{y}} = {\mathtt{13.430\: \!703\: \!308\: \!172\: \!535\: \!8}}\\
\end{array} \right\}$$

Taking only the positive value we then have (approximately)

$${\mathtt{x}} = {{\mathtt{13.430\: \!7}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{x}} = {\mathtt{180.383\: \!702\: \!49}}$$

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If n is an integer then your series is given by ∑_{k=1 to n(}k² - 2)

k      k² - 2    ∑_{k=1 to n(}k² - 2)

1        -1           -1

2          2            1

3          7            8

4          14          22

5          23          45

...

3x - y = 23         (1)

2x + 3y = 8        (2)

Multiply all terms in (1) by 3 

9x - 3y = 69      (3)

Now there are the same number of y's in equation (3) as in equation (2), so add equation (2) to equation (3) term by term:

11 x = 77

so x = 11

Plug this back into equation (1)

3*11 - y = 23  or 33 - y = 23

Subtract 23 from both sides

10 - y = 0

Add y to both sides

10 = y   (or y = 10)

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