Math Is Fun Forum

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

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

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.

Equating those values of y gives

In general

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

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

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

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

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

Yes.  Just make y the subject of the equation.  If you're not sure how to do that here's a step by step.

1)  Multiply out all brackets and multiply by the lowest common denominator to eliminate any fractions.

2) Move terms about so that all the y containing terms are on the left and  everything else is on the right.

3) If there's more than one y term, factorise it out so you have y(......) on the left.

4) Divide by the bracket from the line above so that y is on its own.

eg.

2x/3 + 4(x+y) = x - y + 5

1) clear that bracket and times all by 3

2x + 12x + 12y = 3x - 3y + 15

2) move terms

12y + 3y = 3x - 12x - 2x + 15

3) factorise the y. In this case we can do a lot of simplifying as well.

y(12+ 3) = 15y = -11x + 15

4) y = -11x/15 + 1

Bob

check put x = 15 in the final line so y = -11 + 1 = -10

Substitute these values into what we had at the beginning. If it works then I've probably not made an error.

LHS = 10 + 4 times (15-10) = 10 + 4 times 5 = 30

RHS = 15 --10 +5 = 15 + 10 + 5 = 30

This method of checking can be very useful. The rules of algebra are the same as the rules for numbers so if you choose some numbers to fit an equation at one stage in the working, those same numbers should 'fit' at every line in the working.  If they don't you've found an error with your working.

The forumula for the area of a circle, radius r is

We are told diameters so we need to divide by 2 to get the right radius.  ie 4 and 3

If a ring is a circle with a hole in it you can get the area of the ring by calculating the area of the big circle and subtracting the area of the inside hole.

You're told to take pi as 22/7 which suggests to me there's a simplification in the working involving cancelling that 7.  Let's see:

pi is a common factor so you can do the subtraction and times the answer by pi

Bob

ps.  This example uses substitution into a formula, factorising and cancelling fractions.  Would you like some easier examples of any of these?