Math Is Fun Forum

When I was at school we did tricky multiplications and divisions using log tables so we learnt it earlier than that ie O level stage.  Nowadays it's probably AS.

I can show you the underlying theory if you wish.

Bob

Actually I put the two ys equal and solved for x.  But I could have eliminated the xs and solved for y. That would have worked as well.

If the lines cross then, at the intersection point they both have the same x and y coordinates.  You have two equations so it becomes a simultaneous equations problem. What are the values of x and y that fit both equations simultaneously.  So eliminate one unknown and solve for the other.  As both equations are in the form y = function of x, the quickest way to an answer is to make the two functions of x equal and find the one x that works for both.  Once you have that you can substitite that x value into either equation to get y.  It works whichever equation you choose because that y is the one that fits in both equations.

Many routes to the same answer:

y = 2x -2
y = x/2 -1/2

Subtract the left hand sides and the right hand sides:

0 = 3x/2 - 3/2 so 3x/2 = 3/2 so x=1

substitute in y = 2x -2  ..... y = 2times 1 -2 = 0
substitute in y = x/2 -1/2  ...... y = 1/2 - 1/2 = 0

Make x the subject of each:

x = (y+2)/2
x = 2(y+1/2)
Set these equal

Substitiute in y = 2x - 2  ....... 0 = 2x -  2  ....... x = 1
Substitite in y = x/2 - 1/2  ....... 0 = x/2 - 1/2 ........ x/2 = 1/2   ....... x = 1

Bob

It's all in the name. HCF is highest common factor.  The common factors must be in the intersection because that's what the intersection means.  We want the highest so multiply them together.

Splitting the factors into their primes components helps us to identify the common ones.  You could put {1,2,3,6,7,14,21,42} all in one circle and {1,2,4,7,8,14,28,56} in the other, and then you could spot that 14 is the highest in both sets but it's a lot more work.

Bob

The coordinates of two points before and after the enlargement is enough.  Let's say they are (x1,y1) and (x2,y2) in the first shape and (X1,Y1) and (X2,Y2) in the transformed shape.

Find the equation of the line joining (x1,y1) to (X1,Y1) and also the other similar line. These are the rays but expressed algebraically.

Find where they cross. that's the centre.

You could probably construct a formula for this and then you're independent of a graph entirely.

Bob

If the centre is C and A = (x1,y1) and B = (X1,Y1) then CB/CA gives the scale factor.

Starting with the hexagon divided into 6 equilateral triangles.

Let side = a and apothem = h. Note this is the height of a triangle.

Area of a triangle = half base times height = 0.5 ah.

So area of hexagon = 6 times 0.5 ah = 3ah.  But perimenter = P = 6a so area = 0.5 times 6ah = 0.5 hP.

Using Pythag on a half triangle h = √ [a^2 - (0.5a)^2] = √3 a/2

so area = 3ah = 3a .√3.a/2 = 3√3 a^2 /2.

Bob

The grower bags up the pots and weighs them.  He puts the rounded down amount on the bag so no customer can complain they are being undersold.

Bob

Correct!

Bob

hi paulb203

It looked to me like this should be 'provable' using algebra but I've come unstuck with it.  I don't think I'm properly following what you're suggesting.

If a square has side 'a' then the 'diameter' = a and so the 'radius' = a/2 and the perimeter = 4a

So the ratio of diameter/circumference = a/4a = 1/4.  Ah! Think I've just spotted what to do.

Ratio of circumference/diameter = 4

Then 'area' = 4 x (a/2)^2 = 4 (a^2)/4 = a^2

Bob

I got that too, at first. Then I re-read the question!

The area formula needs the radius not the diameter. So you have to half the numbers given.  Here's another way to think about it.

Take that 8m circle.  You could box it in with a 8m by 8m square and that would have an area of 8x8 = 64.  So the answer must be a lot less than that.

Bob

hi Dombo,

Welcome to the forum.

If you need help then you're at the right place.

What's the problem?

Bob

When you're learning about y = mx + c it's a big help to have equal scales but there going to come a time when you have to deviate from this.

eg y = 1000x + 100.

The MIF function grapher has equal scales.

My ideal would default to equal scales but give the user the option so rescale each axis.

Bob

hi Irene

Algebra is a vast topic, so it's hard for me to know where to begin.

Please give me an example of something you're stuck on?

Bob