Math Is Fun Forum

Normally I'd say yes  but if you do the x cancel out and you're left with an inconsistency.

Just checking Wolfram Alpha

Back again. As suspected WA gives x=0 as the only real solution.

Bob

There are no patterns nor formulas for this, so I think you just have to list and count.

Keep the list: it'll be useful later.

Bob

That looks OK to me. If t=5.1 , abs(t) > 5 and its certainly more than 5 from the origin.  If t = 4.9 it isn't.

Bob

Here's a useful algebraic trick. It will help you to see if your working is right at each step. Choose values for a and b and work out what that makes R. Then use your values of R and b to see if the final value for a is the one you started with.

Normally I wouldn't choose the same for a as for b but your answer didn't look right. I needed a quick check.

I put a = b = 1/2 at the start and got R = 1.

R = 1 and b = 1/2 doesn't give a = 1/2.

I think it's the first R = where the trouble starts. That fraction is the wrong way up. But I think a new approach will get you to the answer more quickly.   Make 1/a the subject and when you've simplified to one fraction invert it for a.

Bob

Each one contains a calculation.  If the result is positive or zero then the absolute sign is redundant; you can simply leave it out. If the result is negative then the result must be made positive (by multiplying by minus 1).  In all of these you can decide whether the result is + or - .

Bob

Yes. If you can get a few right it suggests you've mastered the why.

Bob

I think you've arrived at the right answer.

What is 8.27685 doing in your list?

Bob

Well you could cancel the 2s.   And then no more.

Bob

Linear programming problems are very useful in the real world of business.  They're a way for a company to maximise its profits (or storage space or workforce etc).

I start by making a summary table of the constraints so I can construct the inequalities.  If an equation looks like this

then points on the line will satisfy ax + by = c

points one side of the line will satisfy ax + by < c

and on the other side      ax + by > c

So I start by drawing the line, then pick a random point on one side to see if it fits < or > .  If I can I choose a point with easy to work coordinates ... after all why pick a hard sum to do when an easy one will work just the same. 

Once all the constraint lines are drawn, they should enclose a polygon whose points are those satisfying all.  The thing you've then got to maximise or minimise will be another linear equation except this time we don't know 'c', that's the thing to maximise or minimise.   If you draw a random such line say ax + by = d, then any other line with the same form, ax + by = e, will be parallel.  So tracking across the polygon with parallel lines should enable you to pick out the 'best' point. It will always be a vertex.

You can also have non linear constraints. Don't know whether your syllabus includes those.  Same principles apply but the solution space isn't enclosed by straight lines.  It's then also possible that the 'best' point is on the curve rather than an end point.  I may have gone beyond what you need with this paragraph.

Why not post a past paper question?

Bob

My dear mathxyz,

Please let go of this; it's not getting you anywhere.  At the end of the forum rules it says "Be respectful of Administrators and Moderators as they have to make difficult decisions on behalf of all members."

You have promised more than once to cut down on your posts. I'm happy for you to post requests for help and maths exercises. That's the main point of the forum.  Leave off the other stuff  please and we'll be happy.

If you agree to this, I will expect you to stick to it this time.  If you don't, I'll take out posts.  I am closing this thread. Don't start a new one on this.

Bob

1. won't help

2. yes. Once you have a midpoint calculate the distance from that point to the opposite vertex.  Repeat two more times for the other midpoints.

Bob

This question is on page 41 in my copy of the book.  Mine has pages of introductions before the first lessons and questions.  41-7 = 34. Maybe I'll have to apply a transformation of (my page = (your page) + 34 to find the questions.

The distance formula comes directly from Pythagoras' theorem, so I would end up with the same equation using either.

Let the point be (x,-6)

(x-1)^2 + (-6-2)^2 = 17^2

Solve for x, two solutions.

Bob