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#1 Re: Help Me ! » Algebra » Yesterday 04:33:58



On the y axis the x coordinate is always zero.  The graph cuts the y axis at a point called the intercept.  So just put x=0

You get y=9 I hope.


#2 Re: Ganesh's Puzzles » 10 second questions » 2021-09-22 23:24:35


8108 answer is actually the 8106 answer repeated.



#3 Re: Help Me ! » Algebra » 2021-09-21 19:38:32



Several of these involve straight line graphs.  There are several helpful pages on the main MIF site.

Have a look at these:

And for Q29 (which isn't a straight line):

So what else have we got?

Q23. The rules for algebra are just the rules for arithmetic. So change 'a' to 3 and b to 5 and work out the value of the expression.  Then substitute these same values into each possible answer to see which one is correct.

Q30 is another straight line graph but not given in y = mx + c format.  But you can change it so it is:

20x + 12y = 900

subtract 20 x

12y = -20x + 900

divide by 12

y = -20/12 times x + 900/12

Hope that helps,


#4 Re: Help Me ! » someone help me pls » 2021-09-19 19:54:23


hi new college student

Welcome to the forum.

I think I can help with all three but here we don't just provide model answers to homework questions.  What I'll do is suggest how you can tackle number 1.  When you post back your work on that I'll move on to number 2.  Hope that suits you.

Q1.  There is a method that resembles long division but it's hard to  communicate how it works.  Also you may not be any good at long division anyway in which case it won't help so I'll show you an alternative.

What this question is asking for is this:

where Q is some function of x and R is a number.

This identity has to be true for all x, so putting x=0 will enable you do determine R straight away.

Then you can re-arrange the equation to make Q(x) the subject.

Both f(x) and f(x^7) are GPs (geometric progression). This may help with the simplification.


#5 Re: Help Me ! » Algebra » 2021-09-19 04:38:48


hi jadewest

Q9. For least speed you want the line segment with the lowest upward slope. (so the increase in distance is least for a fixed increase in time).

Q10. The line of best fit seems to have a positive gradient of about 3 and an intercept of -1.  That should make it easy to pick the right equestion.

Q12. The y intercept is when x=0.

Q13. As you have x = for one of the equations it is probably quickest to use substitution.  Put x = 3y + 6 into the other equation to give

2(3y+6) -4y = 8.

Simplify and solve for y.  Then work out x.

Q14. Find x = 110 on the x axis.  Go straight up until you reach the line.  Go horizontally back to the y axis. that's the answer.

Q17 3x - 4y = 9,

You can make x the subject of the equation in two steps. (i) Add 4y to both sides. (ii) Now divide by 3.

But ..... if that is exactly how the question was set then none of these answers is correct.  After step (i) everything has to be divided by 3 so there's a bracket needed.

Hope that helps,


#6 Re: Help Me ! » Composite function : true or false? » 2021-09-18 05:50:33


hi ziabing

Welcome to the forum.

Here's how I tackled this.

I drew a pair of function boxes,; the first showing f; the second g.

Then I tried values of x, starting with x = 1, then 2 then 3 etc.

1 .........f .........0 ......g ......1
2 .........f .........1 ......g ......2
3 .........f .........2 ......g ......3

This is certainly gf (x) = x for x ≥  1

So far so good.

0 ..........f .........1 ......g ......2
-1 .........f .........2 ......g ......3
-2 .........f .........3 ......g ......4
-3 .........f .........4 ......g ......5

That looks good too.

How do you 'prove it'.

For x ≥ 1, f(x) = x-1

So gf(x) = g(x-1) = x-1 + 1 = |x| = x as x is positive in this range.

For x < 1, f(x) = 1 - x

So gf(x) = g(1-x) = |1-x + 1| = |2-x|  As x < 1, 2-x is always positive so the absolute lines are unnecessary, hence gf(x) = 2-x

Hope that helps,


#7 Re: Maths Is Fun - Suggestions and Comments » Inconsistency in the main website and forum names » 2021-09-17 00:42:21


Both lead to the same place.  I think this is simply so that both UK and US new members can find the site even if they get the 'wrong' spelling.

Of course, if you study maths in the UK,  you do lots more as it's a plural. smile


#8 Re: Help Me ! » Multiple Integrals » 2021-09-14 20:06:06


hi 666 bro

I'll have a go at this.  I'm going to go back to basics as it helps me to get things clear in my own head.  Apologies if I'm covering things that you fully understand.

Firstly what is 'ordinary' integration. At school I was told it is the reverse of differentiation but this is not true.  This useful result follows from the fundamental theorem of calculus, but is not the definition.  Sketch a curve and mark on two vertical lines at x=a and x=b.  Integration provides a way to calculate the area between the curve, the x axis, and these two bounding lines.  If you divide up that space into a large set of vertical strips the integration provides a way to sum up the area.  That's why the symbol for integration is a stylised S.

The curve can be written as y = f(x) and the graph is two dimensional.

If a function is of the form z = f(x,y) the  we'd need a three dimensional graph to illustrate it.  That's a tricky thing to do on a flat screen.

Start with a horizontal plane with axes x and y.  For each (x,y) you could compute z = f(x,y) and show this as a vertical line.  Doing the double integration corresponds to calculating the sum of all those vertical lines.  As with the 2D case we need to know the limits of x and y.

I'll write these as a ≤ x ≤ b  , and c ≤ y ≤ d.  Under certain conditions it's ok to keep one variable constant and integrate the other between its limits, and then follow this by integrating the second variable between its limits.  We'd write this like this:

or as

I'll show a simple example first and do the integration both ways.

And now with the integrations reversed:

That's good but not very significant.  It worked so easily (essentially just two integration questions one after the other) because x and y are independent of each other in the function.

I'll now work through a more difficult case where x is a function of y.

The above steps have now been corrected.

Now to switch the order of integration.

When dy is the second integral step the range is 0 ≤ y ≤ 1 and the first step is 1 ≤ x ≤ e^y

This tells us that x and y are connected by the function x = e^y or y = ln(x) where ln is the natural logarithm.

This diagram is the key to making the correct limit swap.


You can see the function y = ln(x).  The upper and lower y limits are shown by the y coordinates of the points (1,0) and (1,1). The lower x limit is shown by the x coordinates of these two points.  The upper x limit has to be a function of y, namely e^y. This much was given by the original double integral that I started with.  Now to get the limits when dy comes before dx.  The dx limits are easier to see.  We start with x  =1.  But where do we stop?  For the function x = e^y, when y = 1, x=e so the top end of the x limit is e. 

So we have 1 ≤ x ≤ e.

Now for the y limits.  The top is easier to see as the highest y gets to is y=1.  But what about the lower limit?  We don't want a number here.  It has to be a function of x, so that, when doing the second step integral, we have a function of x to integrate.  So make it ln(x).

That gives us the y limits as ln(x) ≤ y ≤ 1.  Now I'm ready to do the double integration.


as before.

Finally, why would you ever want to do the same problem twice?  Probably not, but what happens if the integral is very hard or impossible one way, but do-able the other.  Then switching around may be useful.  I don't have a good example of this yet, but I'll have a search and try to find one. If I'm successful I'll add it in a new post. ... LATER EDIT. found one.  Please post if you want to see it.


#9 Re: Maths Is Fun - Suggestions and Comments » Spotted an Error » 2021-09-13 21:55:34


hi Ömer_Bn

Welcome to the forum.

Oh dear. You are right! Well spotted.  I cannot change those pages so I'll message MIF himself.  Fortunately the error is corrected in the next lines so the result stands.

Best wishes,


#10 Re: Exercises » Mathematics » 2021-09-10 00:38:27


hi abbeycity

As far as I know there is no totally algebraic way to solve this.

What I would do is to sketch two graphs, y = 2^x and y = 2x to see where these cross.

You can try this at

In this case they look like they cross at (1,2) and at (2,4);  and it's easy to check by substitution that these are solutions **.  But are they the only ones?

y = 2x is an increasing function, negative when x<0.

y = 2^x is also increasing but never negative.  So we can rule out any negative solutions for x.

2x increases at a steady rate (constant gradient) whereas 2^x gets steeper as x goes up.  So they will never cross again after (2,4)  when the 2^x curve crosses y = 2x with an ever increasing gradient. So  x = 1 and x = 2 are the only solutions.

Does it matter that I spotted the answer without complicated algebraic work?  Well no actually.  If you have shown a solution works and found any others and can prove you've got them all, then that's ok as a way to answer the question.


** You shouldn't assume x= 1 is the answer just from the graph.  The 'correct' answer might be x = 0.9999997.  From a graph alone you only know the answer is roughly 1 as graphs are only as accurate as your ability to draw them (thickness of the pencil; degree of accuracy with the calculator etc)

#11 Re: Help Me ! » Need solutions using Runge Kutta fourth order method » 2021-09-03 21:14:13


hi Rohit K

Welcome to the forum.

It's always useful to know how much the student already knows before embarking on some help.  How much do you know about the second order method? Are you able to post an example of one you have successfully done?


#12 Re: Help Me ! » Topic: Probability: Sample proportion » 2021-08-31 23:28:33


hi Hu9hD0oE

Welcome to the forum.

It's ages since I did this stuff so there's a reliability warning attached to this answer.

I read your post this morning at around 8.00am and I've been thinking about it for the last 4 hours on and off.  I've finally settled on this answer.  Accept it at your own risk.

A 90% confidence interval means you are 90% confident that the true population proportion lies within it.  So there's a 10% chance it won't.

With 40 trials that means you won't get p in the interval 10% of 40 = 4 times.

LATER EDIT: I don't think the next part can be correct.  I'm having a rethink.  Back in another 4 hours smile

I have modified my calculation now.

So let's take 0.1 as the probability that p doesn't lie in the interval. 

It now looks like a binomial with n = 40 and success = 90/100. For once in 40 trials we want 39 'successes' and one 'failure' = 40 * (90/100)^39 * (10/100)^1 and for no successes =  1 * (90/100)^0 * (10/100)^40

I seem to remember you can model this with another normal distribution with n, different p = 90/100 and q = 10/100.   You put in an x (distance from the mean) of 39.5.


How does that sound?


#13 Re: Help Me ! » Area of polygons » 2021-08-31 20:06:08


Yes to both.  Well done!


#14 Re: Help Me ! » Area of polygons » 2021-08-31 19:42:32


Ok number 1:

Assuming that's a rectangle then you can use Pythagoras to work out the height.  Then it's just base x height.

Number 2:

Call the point where the 8 and 10 meet, point A, and the point where 10 and 15 meet point B.  Extend AB until the line meets the 20 line, at point C.  AC splits the polygon into a rectangle and a right angled triangle.  But you need to calculate the lengths AB and BC.

You can get BC using Pythag again and that will allow you to calculate the length and width of the rectangle and the base and height for the triangle.


#15 Re: Help Me ! » Area of polygons » 2021-08-31 19:36:24


hi simonmagusflies

Answers coming up, stay on line please.


#16 Re: Help Me ! » help omg :'( » 2021-08-30 19:23:49



If a = 1/root(5) then

There are two values for k; +1 and -1.  We need to investigate which fits the original expression.


So the value k = -1 is the one we want.


#17 Re: Help Me ! » help omg :'( » 2021-08-29 19:13:15


hi XaiDiv

If you collect the two fractions over a common denominator you'll get:

square this and divide by a squared to get

Then substitute the given value and you're there.


#18 Re: Help Me ! » How to strengthen basics? » 2021-08-24 00:18:27


hi aditsings

Welcome! You're in the right place.  Start by having a look at this site:

You will find lots of basic instruction, well explained, and with many good, interactive things to try.  You'll need to search a bit to find the right pages for you ... but there are good contents pages to help and many cross links between similar topics.

Most pages also have questions to test yourself.

Not everything is yet in place for higher level work.  That's where the forum will be useful.  Just tell us the topics you are struggling with
and we'll do what we can to help.

Also you can post questions that you cannot do and we'll try to help.

Best wishes,


#19 Re: Introductions » New member!!11!!1 :O » 2021-08-22 18:08:52


hi XaiDiv

Welcome to the forum.


#20 Re: Introductions » Just Registered Thanks To Bob And Ganesh! » 2021-08-20 02:58:26


hi Milenne,

Great! Glad that has worked.


#21 Re: Help Me ! » Is it possible to make an account? » 2021-08-19 00:02:38


hi Milenne

Thanks for the reply.

CensorNet is an internet security company that provides protection against all manner of threats.  It looks like it can be purchased by companies and colleges to protect a whole group of users.  If you are trying to create a MathIsfun account whilst logged in at school this might explain the message.  Or maybe your phone company offers this service.

I use Norton 360 and I'm pretty happy with it.  Occasionally we get a post from a person who wants members to click a link that leads to malware.  Norton is so 'keen' to protect me that it won't let me load the page so I can safely remove it.  Luckily I've found a way round this (I won't say what it is because I don't want to help the malware posters); so at the present time I'm confident it's safe to use MIF and to become a member.

So please join up, we're always delighted to have new members. If censornet won't even let you continue the membership process then you'll have to speak to someone to get the restriction lifted.  At a college there'll be a computer services team that will know what to do. If it's your phone company, you'll have to speak to  customer service.

Hope that helps,


#22 Re: Help Me ! » Is it possible to make an account? » 2021-08-18 19:50:32


hi Milenne


It's odd that you are getting an unsafe warning.  I've investigated and cannot, at the moment, find anything that might generate that warning.

Please try again and if you get the warning please tell us:

(1) What security software is giving the warning?

(2) What is the exact wording of the message?

(3) Say what date and time it was when you got the warning.

Thanks, best wishes,


#23 Re: Dark Discussions at Cafe Infinity » Is it possible to upload an avatar for my account ? » 2021-08-14 20:14:09


hi jungleboyx

Welcome to the forum.

The forum server was changed a little while ago.  Some features have improved but the image upload seems to have a bug.  I don't have the means to fix this.  I'll send a message to MIF himself.


#25 Re: Maths Is Fun - Suggestions and Comments » New Members » 2021-08-12 19:21:48


hi jabah013.307

Both sites seem harmless to me so I don't think you need to change anything.  Thanks for the offer.


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