# Math Is Fun Forum

Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

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## #1 Re: Help Me ! » How to find Opposite and Adjacent with only Hypotenuse and an Angle? » Today 08:12:39

That routine for finding a square root was devised by Newton.  My first calculator had only +,- x and divide plus one memory and I used to work out square roots that way in about 30 seconds.

You will find the series expansion here https://www.mathsisfun.com/algebra/taylor-series.html

x has to be in radians not degrees.

There is also a Wiki page that gives the degrees version as well https://en.wikipedia.org/wiki/Sine#Series_definition

You'll have to scroll down to the series definition section.

If you can set that up in css then we have a way forward.  Once you have values for cos and sine there's a relatively simple calculation using +, - and x only to get the coordinates of any rotated point around (0,0), so we'd have to set up the right axes system.

I'll stay subscribed to your thread in the hope you can manage it Bob

## #2 Re: Help Me ! » How to find Opposite and Adjacent with only Hypotenuse and an Angle? » Today 00:19:19

hi Rene,

My short response is I cannot do what you want and don't think it is possible.

Back in my youth we used look up tables to do calculations involving sines and cosines.  Nowadays, we use scientific calculators.  The underlying mathematics for how this works is by the use of 'power series'.  Any sine values can be calculated by adding up terms in an infinite series.  The terms gradually get smaller so, once you achieve a consistent result to a number of decimal places, the summation can stop as no further improvement in accuracy will result.

css does not have either a look up feature nor a power series routine.  It is expected the programmer will use Java or another language.

If two triangles are similar ( have the same angles ) you can calculate additional lengths in one provided you have some lengths in the other.  This doesn't seem to be the case here.

If you limited yourself to certain angles only, then it is possible a solution exists, but I think you want a solution that works at run time and for any angle and that's asking far more from css than its designers built in.

Sorry.

Bob

## #3 Re: Euler Avenue » x ° y = y ° x » Yesterday 20:23:04

Proof that the curve is cut by y = x at (e,e)

differentiate wrt x

If x and y are swapped the equation is unchanged. This means that the function is symmetrical about the line y = x, which means the gradient at the point where the line cuts the curve must be -1

So I can substitute in the above y = x and dy/dx = -1

Bob

## #4 Re: Help Me ! » How to find Opposite and Adjacent with only Hypotenuse and an Angle? » Yesterday 10:03:44

hi renevanderlende,

Great to hear from you again. Let's see if I've understood the problem.

You have a rectangle that you want to rotate through angle θ about one of the corners, let's say corner X.  Before the rotation XY is horizontal.  After the rotation you want to know how far Y has moved horizontally and vertically. You cannot use trig functions.  Is that correct?

Bob

## #5 Re: Euler Avenue » x ° y = y ° x » Yesterday 09:54:22

hi Relentless,

Hhmm. Interesting!

I'm still working on this so I may have more to offer.  Sometimes a graph will help. MIF has two graph plotters; the function grapher will work with equations of the form y = f(x).  The equation grapher only requires an equation connecting x and y.  Here's what I got: I didn't get any negative points.  I think that is just the way the plotter has been set up as we know that negative values are possible.  You can see (2,4) there.  The point where x = y looks like it might be (e,e) but I have yet to prove it.

It would seem that there are values of y for every x and values of x for every y, possibly except where x or y is zero.

I'm hoping to discover more.

Bob

## #6 Re: Puzzles and Games » [Math Challenge]: Prove me wrong. » 2021-10-07 05:58:02

Ok. think on this:

Forum rules

No Personal Attacks or Put-Downs. This is a type of bullying, and just makes you look insecure.

This is not a place to be mean to others and these posts will not be tolerated. Light banter or constructive criticism can be allowed if it is polite and friendly. Remember, other people have feelings too. "Those who give respect shall receive it."

Possible Actions: At first you will be gently warned or have your message edited or deleted. More serious cases may result in banning or other measures.

Please regard this as a gentle warning.

## #7 Re: Help Me ! » Algebra 2 (functions) » 2021-10-06 19:53:08

hi Ava,

Welcome to the forum.

That's a lot of similar questions, so, hopefully, once you've done one ok, you'll manage the rest on your own.

Think of a function as a box with a number that goes into the box and another number comes out.

Q1. f(x) = 15x – 12 and g(x) = -15x2 + 14x - 10
find g(f(7))

This means put 7 through an 'f' box and then the answer through a 'g' box.

So do 15 times 7 - 12; get an answer and apply the g function to that answer.

Q4. g[f(x)] if g(x) = x2 and f(x) = x + 3.

Same as with numbers but you have to maintain the algebra in going from step one to step two.

So if you put x into the 'f' box, x + 3 comes out. Now put that into the 'g' box (which squares the input)
and (x + 3)^2 will come out.

For some you might need to  do some algebraic simplifying, but I don't think this one needs that.

As any number can have 3 added and any number can be squared the domain can be all real numbers.

Please have a try at these and, if you want, you may post your answers back for checking.

Bob

## #8 Re: Introductions » Hello » 2021-10-04 06:33:49

hi Rusty,

Welcome to the forum.

All sorts of topic areas are covered in the main teaching site.  https://www.mathsisfun.com/

And when that's not enough you can ask here. Bob

## #9 Re: Help Me ! » Algebra » 2021-09-29 19:00:13

Answer A looks good to me. Bob

## #10 Re: Help Me ! » Algebra » 2021-09-26 04:33:58

hi

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.

Bob

Whoops!

Bob

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

hi

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

Have a look at these:

https://www.mathsisfun.com/equation_of_line.html

https://www.mathsisfun.com/data/straight_line_graph.html

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

https://www.mathsisfun.com/data/function-grapher.php

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,

Bob

## #13 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.

Bob

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

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,

Bob

## #15 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,

Bob

## #16 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. Bob

## #17 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.

Bob

## #18 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,

Bob

## #19 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 https://www.mathsisfun.com/data/function-grapher.php

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.

Bob

** 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)

## #20 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?

Bob

## #21 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 LATER LATER EDIT:

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.

and

How does that sound?

Bob

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

Yes to both.  Well done!

B

## #23 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.

Bob

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

hi simonmagusflies

Bob

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

so

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

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

Bob