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hi Glenys,

Welcome to the forum.

Yes, you've come to the right place. I have taught maths in Uk secondary schools for 37 years so I ought to be able to help.

Here's a start.

Maths is full of equations for all sorts of things. For example the formula for converting degrees fahrenheit into celsius is this

I've used Latex to make that appear 'nicely'. I typed C = (F-32) \times \frac{5}{9} and then boxed it in with square brackets math and square brackets /math

If you click on any piece of Latex you'll see the underlying 'code' that made the piece.

Descartes invented the x-y graphing idea so you could make a graph of that formula by using y and x instead:

If you go to this page

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

and change the equation to (x-32)*5/9 you'll get the graph for changing F into C. [note that the grapher 'understands' * for multiply and / for divide]

Don't panic when no graph appears. It doesn't show with the basic default scale. You'll have to adjust the scale using the zoom (drag right) and you can click and drag to see other bits of the graph.

Equations where x occurs like that give rise to straight line graphs.

If an x squared term occurs (usually written as x^2 on this forum) then the equation is called a quadratic, and the graph is always a curve.

While you're looking at the grapher page try this equation:

and you will see a typical quadratic graph.

There's several examples of quadratics here:

https://www.mathsisfun.com/algebra/quad … world.html

Hopefully that will give you a starting point. Post back when you want more

Bob

hi mrpace

Bob

hi cooperblu

Welcome to the forum.

This same question has been asked before here:

http://www.mathisfunforum.com/viewtopic.php?id=25850

Bob

hi

I haven't forgotten you. I'm working on a couple of approaches which may help. Is it possible for you to post the working you tried ... don't worry if it's muddly ... I'll try to work it through.

Bob

hi blob123

Welcome to the forum.

One caution. I'm a bit shaky on questions like this so I may be wrong but here's how I'd tackle these.

Make the question easier by simplifying the numbers.

Let's say you have just 6 teams and you're trying to make up three matches. I'll call them C1 C2 C3 C4 M1 and M2

How many ways can we select the matches?

Pick any team. There are 5 opponents to choose from. Pick another team. There are 3 opponents to choose from. Pick a final team. There is just 1 opponent to 'choose' from.

So that makes 5 x 3 x 1 = 15 possible selections.

EDIT: I've had a further think and the next bit isn't correct yet. I still think it's ok to try and simplify the problem. I'm working on a modification that will yield the correct version. Hope to get there later today. LATER EDIT. This is now corrected.

Now require that just two Cs are together. There are so few possibilities I can easily list them:

{C1C2 C3M1 C4M2} {C1C2 C3M2 C4M1} {C1C3 C2M1 C4M2} {C1C3 C2M2 C4M1} {C1C4 C2M1 C3M2} {C1C4 C2M2 C3M1} {C1M1 C2C3 C4M2} {C1M1 C2C4 C3M2} {C1M1 C3C4 C3M2} {C1M2 C2C3 C4M1} {C1M1 C2C4 C3M1} {C1M2 C3C4 C2M1}

So I have 12 possibilities making a probability of having just2 Cs of 12/15 = 4/5

Is there a way to get this without listing all the possibilities?

Start with a C (4 choices) Choose another (3 choices) Divide by 2 as I'm counting eg C1C2 and C2C1.

The remaining Cs must each be paired with an M; there's just two ways to do this. (eg C3M1 or C3M2 ... last pair is fixed).

So we have 4 x 3 / 2 x 2 = 12.

That approach can now be scaled up for the 20 teams.

Bob

hi mrpace

You need to create a mathematical model to 'describe' this situation. With only two parties the binomial distribution might work.

Probability in this 'constituency' of a voter supporting A is 0.65 (p) and for B is 0.35 (q).

Now you have to assume that these probabilities will remain for the rest of the country and that's where this comes unstuck.

Expected number of votes for A = np = 1200 000 x 0.65 = 780 000

Standard deviation is √ (npq) = 522.

You can use the normal distribution approximation for binomial results, which I have tried. I don't think it's worth showing the calculation because it makes it virtually certain that A will win overall.

The problem is with the underlined assumption. Think about it. 0.65 is pretty high and if that's the probability in every constituency then A will keep winning. In the UK that doesn't happen. There are big regional variations between support for the parties so the 'pundits' know they cannot assume general results from a single announcement. The pre-election opinion polls carefully select their samples by spreading across areas with so many people of each gender and age group.

To do what you want, you'd have to know the probabilities for each constituency; then you could calculate on an area by area basis.

The underlying theory is here: https://www.mathsisfun.com/data/standar … ution.html

The method works ok for bags of sugar if it's ok to assume consistency during the production run. Voters are more complicated.

In the UK they do an 'exit poll' asking a sample of people how they voted after they have voted. Providing the sample is across all areas like the opinion polls this does usually give a good prediction of the final outcome.

Bob

hi 666bro,

I'll use the example from above.

We know there's a turning point at √(1/3) ≈ 0.57735

Investigate the gradient just left and just right of the turning point. I've chosen x = 0 and x = 1 to make the calculation easy.

The top line in the table shows the x values chosen and the second line shows the gradient at those points.

Below that I've sketched those gradients and the sketch shows it's a minimum.

Bob

hi Pema,

Welcome to the forum.

Thanks to zetafunc for the initial hint.

You might also find the following useful:

w^2 + w + 1 = 0 => w + 1 = - w^2

You can prove this equation either by substituting the complex values for each root (the working is identical) or like this:

Let w^2 + w + 1 = z ; then w^3 + w^2 + w = wz

As w^3 = 1 we end up with z = wz. So either z = 0 or w = 1. As we know that second isn't the case z must be 0.

Bob

hi juney06

Welcome to the forum. Thanks for posting your answers to the earlier parts of the problem. It's a great help in knowing what you do understand about the problem.

I started with a diagram and marked on the height of the cliff and the initial velocity upwards. This is 'up' but the cliff is down and so is the direction of gravity. So you need to decide at the start which way you will take as the positive direction and then use + or - correctly for the info given.

You've got the right formula to use but the signs aren't consistent. If you take 'up' as positive then u = 24.5, g = - 9.8, and s = - 117.6 (that's taking the cliff top as zero).

So the quadratic comes out as

If you take 'down' as positive then it is

As you can see these amount to the same thing so it doesn't matter which you use; but be consistent.

I tried this and got 8 seconds but not with your quadratic. I've just tried to solve your version with a quadratic solver and the square root part comes out negative so it cannot be evaluated. So did you make another sign error which 'cancelled out' the first?

For the last part the ball is to be 49m above the ground. But we are both measuring from the cliff top so you need to work out the distance down from the top to the 49 position ie 117.6 - 49. Then you can use that as your new 's' value to once again get a quadratic.

Hope that helps,

Bob

hi 666bro,

It's great to hear that things are going well for you.

I'll illustrate what's happening with an example.

I've chosen the function

When we differentiate we get

and second derivative

Here are the graphs of (black) the function; (red) the first derivative; (green) the second derivative.

I've put the three graphs one after the other, lined up in the x direction but one on top of the other in the y direction. For this reason don't worry about the numbering on the axes; you don't need it. But the changes in gradient for the function are aligned correctly with the first and second derivatives.

The function is a cubic. When you differentiate it you get a quadratic. The quadratic has two zeros and these line up with the turning points on the cubic. When you differentiate again you are finding the gradient function of dy/dx, and you get a linear function.

On the (black) function graph you can see the two turning points; the first a maximum; the second a minimum; but let's say we don't know this yet. Just to the left of a maximum the function has positive gradient; just to the right it is negative. At the turning point the gradient is zero of course.

You can see this on the red graph. Just to the left of the function's turning point the red graph is above the axis so it's values are positive. The red graph crosses its axis at the turning point and after that it has negative values. In a similar way the next turning point (a minimum) has gradient negative to the left of the turning point; zero at the turning point; then positive.

So you can tell if a turning point is a maximum by looking to see if the first derivative goes from positive, through zero, to negative. And a minimum will have first derivative that goes from negative, through zero, to positive.

One way to investigate this is to draw the graphs; another is to calculate gradients just left and right of the turning point. But the second derivative gives a quick way to find out without needing to do any graph drawing. Here's how:

Function has a maximum. Gradient function goes from positive, through zero to negative. So its gradient function must be negative at that point as dy/dx has a reducing gradient.

Function has a minimum. Gradient function goes from negative, through zero, to positive. So its gradient function must be positive at that point as dy/dx has an increasing gradient.

In my example

so the turning points are at x = - √ (1/3) and + √ (1/3) At x = - √ (1/3) the second derivative is negative => this turning point is a minimum.At x = + √ (1/3) the second derivative is postive => this turning point is a maximum.

Note: I don't even have to do the full calculation here; I just need to know if the second derivative is positive or negative at each turning point; so it's a quick way to tell.

Bob

hi Double

Welcome to the forum.

Sketch the graph and draw two horizontal lines close together from the y axis to the curve, so you have drawn a typical strip. This strip has area x across and dy width. If you were asked for the area of the part of the curve between y = 0 and y = 1 you would add up the strips like this:

You cannot integrate a function of x with respect to y, so you need to substitute with

The resulting function is directly integrable.

But you ask for a volume, so I'll assume this means that the area is rotated around the y axis to create a solid 'volume of revolution'.

Go back to the original strip and imagine it rotates around the y axis so as to generate a thin disc. The area of this disc is π x^2 ( ie. pi radius squared ) and so its volume is π x^2 .dy Here the new integral:

Once again you need to replace the function of x with one involving y ....... x^2 = 1 - y

So we then end up with this integral:

Can you finish from here?

Bob

hi Hugo28036

Ok. So what is the algorithm?

Bob

hi sarahparker

New members who want to share a love of maths are always welcome. The forum is funded from sponsors on the main site, mathsisfun.com, and it is a condition of membership that posts must not contain advertising. I have therefore removed your website link as that looks to me like advertising.

Bob Bundy

hi 666 bro

I'm glad that your work on matrices has gone well.

The derivative of a function is the slope (gradient) of the graph at points along the curve.

example.

y = x^2

Just by looking at the graph we can see that at x = 0 the gradient is zero; that it rises up through positive values and has matching but negative values when x < 0

Differentiating the function we get dy/dx = 2x and this has the right gradient properties. When x = 0 2x is zero. At say x = 3 the gradient is 6 and at x = -3 it is -6.

There's lots on this on the MIF page https://www.mathsisfun.com/calculus/der … ction.html and there are links to other pages on calculus from there.

Bob

hi isabella.bueso

Welcome to the forum.

I'm happy to help students where they are having difficulty with a topic in maths but, if you read the page about what to do before you post you'll understand that we don't just do your homework for you.

This worksheet has been posted before. The company that makes these worksheets doesn't like it's copyright being abused by postings on the public forums.

There's no diagrams. What chance do I have helping with questions that are meaningless without the diagram.

When I am teaching a class, I know the past history so I know what they already know and where they might struggle, so I can target my teaching approach more usefully.

If you really cannot do any of this sheet then you need to get back to your teachers and ask for more lessons.

So, if you want my help, please say

(1) What can you do on this topic.

(2) Show what you have tried already.

(3) Be specific about what you are finding difficult.

Bob

hi George,

There is an automatic word swap 'censor' that changes certain words that are thought to be rude in some way.

Anyone who wants to click the link can first save it and then substitute back the right word instead of the word math.

It's a nuisance but it's there to protect the site from abusers.

Edit: Just tried that and here's your picture:

Bob

hi Someshwar Tripathi

Welcome to the forum.

This picture shows

You'll find lots more help here:

https://www.mathsisfun.com/improper-fractions.html

Bob

hi johnny453

Welcome to the forum.

No diagrams? And no explanation of what this is about. Sorry but it'll be tough to help under the circumstances.

I'm going to assume this is an exercise on enlargements and scale factors. Here's an example of a quadrilateral that has been enlarged with a scale factor of x2.

AB is 4cm long and CD is 8cm long. The other lengths have doubled in size similarly.

If you don't know the scale factor you can work it out like this:

I not sure what 'prop' means in these questions. In Q8 one possible answer is not prop, so I'm guessing that some questions are deliberately not proper enlargements. How can you tell?

You have to check every enlarged length. If the sf is always the same then you can safely conclude that you have a proper enlargement with that sf. If some sf answers come out differently from others, then that question is not a proper enlargement.

Hope that helps,

Bob

hi Otsdarva

My method is

From this you get roots of -4 and 0, and hence the x coordinate of the vertex is -2, leading to a y coordinate of +6.

Looking at your method:

Taking out the -3/2 is OK but you need to maintain the factor as you complete the square, otherwise you are changing the quadratic.

Bob

hi Phro,

As you've used less than signs rather than less than or equal, your set of values is very small indeed.

Mathegocart

If I show you my proof you'll understand this, I think.

Bob

hi Knewlogik

How interesting! But can you prove it always works?

Bob

Hi,

You are very welcome. It's been a pleasure to help someone who is so polite and has such a good grasp of the topic. I have always had a great regard for any pupil who really thinks it through and doesn't just go for the 'expected' answer. Hope the rest of your studies go well but do post again, either for help or just so we know how it's going.

Bob

hi sybil8464

Your answer for Q2 77.78 is correct. You need 14+ out of 18. 14/18 is 77.7777.. % so choosing 77.78 does that.

For Q7 you need 18+/21 . 18/21 is 85.71428..... so strictly 85.71 is not enough. But it's the highest answer so it'll have to do. These aren't from Compu-High by any chance are they? Because they use multi-choice and computer marking you cannot say none of these actually work, so try 85.71 and if you can point out to your tutor that it doesn't really work maybe they'll realise they need to take more care setting their questions.

For Q10 you need 15+/18. 15/18 = 83.3333 % so I can see why your sister suggests that, but once again it's not strictly enough to get past 50% of the votes. So i like your thinking in suggesting answer 4. However, I suspect the person who set the questions hadn't thought this through properly because of rounding. I'm not sure how you answer this. If you can write an explanation for your answer then you deserve extra credit but are their tutors clever enough to spot this? Please post back what you submit and how it is marked.

Best wishes,

Bob

hi sybil8464

Welcome to the forum.

From the way you have described this, either answer is correct. The question setter needed to say clockwise or anticlockwise. Normally one would give the lower value ( 150 ) but, unless the wording makes it clear, 210 would also be good. In some areas of mathematics anticlockwise is the norm, so I'd suggest you say 150.

Hope that is good,

Bob

hi PMH

Go up one from k1, and test the block from there to F, using no more than k-1 drops.

k2 = k1 + 1

Shouldn't that be k2 = k1 -1 ?

This generates the series k1, k1+1, k1+2, ...

which has to sum to F.

Sum: k*(k+1)/2 + k*k*(k+1)/2

= (k+1)*(k+1)/2

so F = (k+1)^2/2

so k = sqrt(2F)

I think that is Sum = k(k+1)^2 over 2

Bob