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It is imprecise. As it stands the locus should include points inside the stadium. That covers the point you're making.

If they only wanted the lines and semicircles it should specify the shortest distance.

Bob

Did you consider the direction of your infinite line? I have received a hyperwave transmission from Alpha Centauri complaining about an infinite line that keeps intersecting their planet (every 24 hours) and, crucially, is interfering with their reception of House* on the interstellar carrier transmitter/receiver. They believe the line originated in your area.

* They are big fans of Hugh Laurie.

It may only be imaginary to you, but, in interstellar carrier transmission/reception, imaginary lines play a significant role. Indeed, without complex numbers it's doubtful such things would be possible

Bob

Using compasses to construct the perpendicular bisector is the correct method. (You could also measure for the midpoint and use a protractor for the perpendicular line but see note below about awarding marks for the construction)

In theory the line extends to infinity in both directions. Clearly, in practice, you have to end the line somewhere. A marker would be looking for your construction so don't erase anything. To show you understand the locus doesn't end where your arcs intersected extend the line beyond those points.

If you really want to be mathematically pedantic you could add a note that your locus extends to infinity.

Bob

Hhmmm. Very primitive. My dial doesn't need batteries and will function for thousands of years. Only snag; the sun has to be shining.

Bob

Things are going well for me.

Some years ago I built a sundial in the back garden but the post had gradually sunk into the soil, but unevenly, so the clock face was no longer horizontal. I took a major decision to lift it and reset it in post cement. But I lost the orientation on the pole star so last night I was out back at midnight finalising the adjustment.

I built it with two circular dials. The top one is fixed and points along the polar axis. The lower one can be rotated. Half the circle has the hour angles for GMT (6am to 6pm with 12noon in the middle) and the other for BST (7am to 7pm with 1pm in the middle). That way I can change it around when we switch to 'summer time' and back for the winter. I have a graph for the equation of time with longitude correction included which yields a small angle correction so a fine adjustment can be made to the clock face to get a really accurate time throughout the year.

Bob

hi Frank smith

Welcome to the forum.

It's a good question.

If you look here: https://www.mathsisfun.com/geometry/cir … ctors.html

you'll see a way to establish the formula by dividing a circle into sectors and letting the number of sectors become infinite.

But there's still a question because this method uses the formula for the circumference, so is there a proof for this too?

I cannot find one on the MIF site; I'm sure you could 'google' it.

I've got a method that works out the perimeter of a regular polygon and lets the number of sides tend to infinity so the polygon becomes a circle.

It uses sin(x) tends to x as x tends to zero (x in radians). There a justification* for that here:

https://www.mathsisfun.com/geometry/radians.html

* The method I learnt at school uses the formula for circumference so it isn't suitable here as it creates a circular argument (no pun intended).

Bob

hi Agnishom,

Great to hear from you again.

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Bob

You would need to know what 'h' represents. 140 < h ≤ 150 implies h can have any value above 140 with an inclusive upper bound of 150. So h could be 140.000000000001

If you draw a number line with those endpoints, 145 is half way along (and having one excluded endpoint and one included does not alter that).

If only the numbers {141,142,143,144,145,146,147,148,149,150} are under consideration such as "What is the median of these ten numbers: 141,142,143,144,145,146,147,148,149,150" then 145.5 is correct.

So median and middle number may not be the same thing.

Bob

?? It's the condition, so what more are you expecting? Did you want the single solution too? It wasn't asked for was it?

Bob

Bob

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

Yes if you use logs.

You can have logs in base 10 or the natural log base (e) but both logs have to use the same.

Bob

Using 1.08 I calculated n as 3.99999999 in Excel so that seems ok.

Why do we equate the two values of x?

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

Equating those values of y gives

In general

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

To make a Venn diagram, draw one circle for each number. Put the prime factors inside the circles so that common primes are in the intersecting part of the diagram.

The union gives the LCM.

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.

I think it's odd that the question refers to small bars and large bars. Does that mean there are two sizes of bar? I think the question just means bars that are put in small packs and same size bars that are piut in large packs. Altogether a very badly worded and thought out question. It would be easy to change the numbers a bit so that only whole packs are sold and still test the same skills.

Bob

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

My working is a little different but comes to the same answer, 50 small bars leading to 12.5 packs. you can get the answer even though the answer makes no sense. I've re-checked this several ways and still get to the same conclusion.

Odd

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

You're talking about a regular hexagon I think; so the answer is yes. You can divide the hexagon into 6 equilateral triangles each with side = the same as for the hexagon itself. Let's say side = a. Then perimeter = 6a and diagonal = 2a.

Bob

Correct!

Bob

5x= 6 means x = 1.2

An equation that gives rise to a line is why the word linear is used.

It's the absence of any powers eg x^2 that allows this to happen.

This usage has spilt over into equations generally so in your examples linear is used because there's no powers .

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

sqi = the ratio of the ‘diameter’ of the square to the square’s ‘circumference’

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