Thanks so much - i've got the answer now :)

If 2^2n was simply replaced with "x" you'd have

x + 4x = ...

Just done my further maths exams in school a couple of weeks ago and we did AQA, which I found really quite easy and enjoyable. Wasn't time consuming and gives a good head start to AS. Edexcel FM is a lot harder I believe and covers more content. It is also an IGCSE compared to AQA's level 2 certificate. Use Corbett maths playlist if you do the AQA spec

And so total should be

1800+380=2180

:)

I've transferred to an International School abroad that uses P1-P4 instead of C1-C4. Is there a website available where I can study P1-P4 with self-paced learning and availability to contact the tutors?

Any website would be helpful as long as it helps me prepare for the exams especially with P1-P2 as I'll be going into Year 13 and have to take P1 and P2 as I took C1 and C2.

Its worth noting you're just asking about a simple linear scaling of x. So if you have a very simple function

y = x

and you consider a prior linear scaling of x -> kx, then the gradient (with respect to x) is k. The larger k is, the more sensitive (larger gradient) y is to changes in x.

For a more general function

y = f(x)

with a prior linear scaling x -> kx, you can use the same argument as the gradient is defined in terms of the local tangent so the gradient of the composed function (linear scaling then f()) is

k*f'(kx)

Obviously, the exponential you mention is one example. Another simple one is the gradient of sin(kx)

k*cos(kx)

Also, the derivative of ln(kx) is

k*(1/kx) = 1/x

as you'd expect. And as has been mentioned one or twice, its a simple example of the chain rule.

Past Papers all the way

Although there are similarities to an inverse function, we're not finding an inverse function as such; indeed, there isn't one.

I am assumng the domain is the whole of R, bar where the denominator is 0, which is what is normally assumed for these types of questions.

y = 1E-11x5 - 2E-08x4 9E-06x3 - 0.0017x2 0.1353x - 3.2776 ?

You've not posted the question, but I guess its something like the value of n for which the sum of the geometric sequence is within 0.001 of the final value. n=14 would be too small to satisfy the inequality, whereas n=15 is the first integer that satisfes the inequality, whatever that inequality actually is.

You'd lose marks if you incorrectly rounded n down to 14.

PM if you would like a copy of the paper

Just to add that if you want to continue in this thread that's fine too.

If you say where you're stuck (we can look up the paper, so you should be able to describe without needing to typeset lots of maths), someone will help you through it.

Hi i think that it is recommended u hv it but not required

Good luck with the future

Duplicate thread.

See https://www.thestudentroom.co.uk/sho....php?t=7040093

(a) Find a graph with degree sequence 4, 4, 3, 3, 2, 2, 2, 2.

[3 Marks]

(b) Show that any graph with this degree sequence is planar.

[4 Marks]

(c) Is it true in general that if a graph is planar, then every graph with the same degree sequence

is also planar? If true, prove the statement, if false, find a counterexample.

Well induction is the suggestion.

Do the base case I'd suggest all graphs up to 2 vertices.

Then looking at a directed complete graph of n+1 vertices, delete one vertex and its associated edges. The resulting subgraph has n vertices, and so has a Hamiltonian path. Now add the extra vertex and show how the path can be adjusted to get a Hamiltonian path on n+1 vertices. There are three cases to deal with.

Guess you both need to read it.

https://www.thestudentroom.co.uk/sho....php?t=4919248

Remember to post your attempt with the original question.

Calculators dont integrate the formal equation, they only numerically integrate using an approximation, similar to something like simpsons rule.

So if a paper has determined a question because it gives a convenient fraction based on the specific equation and boundaries the calculator may give a very very close answer to an extreme number of decimal places but its likely not the same as the exact answer (especially if the mark scheme answer is a constant multiplied by root2, pi, e etc.)

My list was not exhaustive and that's all i could think of from the top of my head. These are normally only 1 mark questions anyway so only 1 is needed. Honestly the fact that he could just have googled it and got an answer in like 5 seconds or use a textbook and get the same result illustrates they have no lateral thinking anyway so i don't have much hope for them in any regard. I answered it cos I wanted to see what i remembered lol

The heights, F cm, of an adult female population are normally distributed with mean 162 cm

and standard deviation 7.5 cm.

When two solids are similar, and the length of the second shape is k times the length of the first shape, by what factor must the area and the volume increase? (Think about how it is enlarged by k in more than one axis). Use this to answer your question. If the area increases by a certain factor, then what must the length or volume increase by?

Usually better to start your own thread in those circumstances - otherwise people start quoting TSR users who've departed many years ago :D

Also, it's not appropriate to post full solutions - check out the sticky post at the top of the forum for posting guidelines :)

does anyone happen to have a digital copy of the paper or markscheme?

oh he spittin

I would say it is easier to not use a tree diagram. There are 2 orders in which she can pick the cards: 1 then 2, or 2 then 1. Lets do 1 then 2 first. There are 3 ones out of 5 cards, so probability is 3/5, now that she chooses a card, there are 4 cards remaining. Out of those 4, two are 2s, so the probability of picking a 2 is 2/4. Now we multiply them(3/5 x 2/4), which is 3/10. Now the same can be done again but this time 2 first then 1. There are two 2s out of 5 cards, so probability is 2/5. There are 4 cards remaining, 3 of them are 1s, so probability is 3/4. Multiply them together(2/5 x 3/4), which is 3/10. Now we add, (3/10 3/10), which is 3/5.

bii) would be pythagoras as you describe.

This is the debate section. Not the please-do-my-homework-and-thinking-for-me section.

At least tell us what effort you've made so far.

maths section?

"Complete Mathematics for Cambridge IGCSE Student Book (Extended)"

Fifth Edition Published by Oxford, author isDavid Rayner

A question about what exactly is in the book, does the book cover both the core and extended syllabus or only the extended syllabus?

If I want the complete extended syllabus, do I need to buy both of the books CORE and EXTENDED or is it enough to buy the extended book. The reason I ask, recently I bought a hodder book for IB maths HL. Turns out I also needed to buy the IB maths SL book as the HL book was for only the HL part of syllabus.

I am not in the UK and would appreciate if someone could kindly advise.

a link to the Extended book here:

https://www.amazon.co.uk/Complete-Ma...Q19N2ETZZNFJEP

a link to the CORE book here:

https://www.amazon.co.uk/Complete-Ma...3224274&sr=8-3

I have been using the thr

sin^(m-1)x/cos^(n+1)x. cos^2x? Is this OK?

Ask your teacher

No idea but my English one was 2017

Try plotting it on desmos with different values of x :) https://www.desmos.com/calculator

Think about what you know about ln functions, when is it defined?

Thank you!

no, it's the integral of a positive function which must be positive, you will have made a mistake somewhere (feel free to post the working)

Depends if your referring to 1/4(x) or 1/(4x). The first one is the same, the second is not

The maximum is achieved when the bag with the smallest number of coins has just dimes, the next smallest bag has just quarters, the third smallest bag has just 50 cent pieces, and every other bag has just dollar coins.

https://www.mathsgenie.co.uk/gcse.html

Year 11 and year 13 students have mostly finished their assessments now so most tutors will have a few spaces opening up round about now.

Fair enough. Maye I'm over-thinking.

The data is side effect rates from certain drugs. I'm looking at 3 particular side effects at the moment, though will have data on more later. There are 5-8 different studies, depending on the particular side effect, with the ranges being 4-30%, 37-69%, and 34-71% for the three, to give an idea of distribution between studies.

N ranges from 54 to 410.

Quote:

R makes life a lot easier if you are able to use it!

I am attending a 1 day course on how to use R, but not for a couple months yet!

как бихте отговорили на приемния изпит,

моля

A line is a function (curve)?

Wait sorry I actually do have a question.

q) give a reason to support the use of a histogram to present this data

a) the data given is a continuous variable and the data is given in a grouped frequency table

why is it relevant for a histogram that the data is a continuous variable and is given in a grouped freq table? I don't understand..🤷

In a class 2/5of the students have blue eyes after 25 students in the class how many have blue eyes

With maths practice is key so try exam questions on maths genie. If you’re looking for videos exam solutions is good but they’re a lot longer than free science lessons. For something that’s just as quick primrose kitten has a video that goes over the whole course.

I believe the answer is D.CalculationIn the nominal ride the passing point is the library however with a twenty minute delay the passing point is 3km earlier.That means the difference in the twenty minutes on THIS ride is 3km in twenty minutes or 9km an hour. This is the excess speed of Charles which we will use later.This plus Claire’s initial walking speed is 15km/h however this is only part of the equation and therefore the answer is not 15.Claire would only walk 2km in the twenty minutes not the 3km difference meaning that Charles must be going half as fast as Claire again (3/2=1.5). As such we need to multiply Claire’s speed by 1.5.Difference of Claire’s speed1.5x6=9kmhPlus the extra speed of a Charles (previously determined 9km/h) 18km/h answer D

Hi guys, please help with this if you know

suppose that a follow-up survey with the same questions was conducted in June 2022. Describe, with the aid of an appropriate diagram labelled with variables, how you could test whether the same factors existed at this later date.am not sure what to use here ?

You have to do integration by parts with 1 and lnx

Similar question to previous, it looks similar to your recent, previous posts which you said was assessed. If so you shouldn't be posting assessed questions / solutions.

However, not sure of your formula or units. You seem to be doing radial acceleration / uniform motion.

bump :/

If the question is correct, then note that so .

So you cannot model this and find the length of the missing sides using right-angled triangles in the way you have (as none of the angles of the triangles can be 90 degrees).

Also, as mentioned earlier, you are mixing up the hypotenuse with opposite.

As for how to answer this question, what I tend to do is draw the an Argand diagram of the complex value then (using the appropriate angle) I use the method you used.

My diagram would look similar to the image below:

HINT: Find the angle in red (in the image), then use the method you used to find the length of the sides.

FURTHER HINT: There are two solutions to (where ), use the method above to see if you can find the other solutions.

Finally, can you upload an image of the answers at all (to see if the answers correspond with the question)?

Casio fx-991EX. Costs about £30. Has all the statistical distribution functions you'll need, numerically solves equations and integrals - good for checking your own work.

It probably won't, but its always good to review and learn some basic geometry.

For questions like this, they want you to spot the base * perp height / 2. Its quicker to work out the area like this, than to calculate the lengths of the two sides which are not parallel to the axes.

Note the other question was 3 marks. You'd aleady found the + C y-axis intersection points, so a couple of easy lines to find the area. Nothing more is required for 3 marks.

Is this another one you're being assessed on? If so, you shouldn't be posting questions/solutions for checking.

A couple of hints anyway,

* plot the x-y-t curves in desmos. its easy to spot/validate where the gradient is zero etc

* are you working in degrees or radians. I realize your answer is the former, but you're differentiating so ...

I've just finished this and passed with Open Awards qualification through Think Online learning. However, I did use practice papers from different awarding bodies and found them to be similar.

Note you've calculated the x-axis intersect, rather than the y-axis for the normal (point B)? So calculating the area should be simple.

Each of those values of a has two solutions, one of which is x=4.

For one value of a, x=4 is the largest solution. For the other value of a, x=4 is the smallest solution. The question asks for the second case and wants the other value of x which is a solution. Obviously, the solution must satisfy x>4.

Choose one of these:

Top Homework Helper

Khan Academy

Study Geek

Tutor.com

Chegg

Forum guidelines are that you post your attempts first. There are lots of examples for the basic log rules like

a*log(x) = log(x^a)

log(x) + log(y) = log(xy)

https://www.mathsisfun.com/algebra/e...ogarithms.html

Can someone explain to what will happen when an extra pipe is added.

Use this website: https://www.drfrostmaths.com/sow.php...2017&term=Main

There are worked examples and it's the one I recommend to my students.

Its easy with questions like this to just view it as an algebraic slog. As well as validating your answer by subbing it into the original equation you could note that each side of the original equation could represented as a line

y = mx + c

and the equation represents the crossing point (where the two lines have the same x,y point). If you approximate each side (take each number to the nearest integer) you'd have something like

2x-4 = x

so the answer will be roughly 4. The left hand side starts (when x=0) 4 underneath the right hand side but the gradients have a difference of 1, so you need 4 x increments for them to meet.

Both validating your calculated answer and approximating/visualizing the original equation can help find errors.

What are your thoughts so far?

If you don't have any idea where to start, look at what happens for smaller powers and see if that helps you form a hypothesis.

its really just about understanding trigonometry & A-level mechanics.

Their is no trick, you'll just need to understand the kinematics and how forces act on objects, for the most part the Sin & cos part is more just applying GCSE trigonometry to the numerical values dictated in the kinematics question.

fajl.

Let Y=x^3/4

Use this substitution to easily expand. Then simplify it through that. Remember to change back to x.

(3Y-1/Y)^2

.....

1/2 isn't a root.

guessing angle x is the angle at the top of the triangle.

Angles in a triangle add to 180 degrees

so add 65 and 45, this gives you 110 degrees

180 minus 110 = 70 so

angle x = 70 degrees

thank you for your help :)

Thank you!

8sin30 is the component of the mass' weight parallel to the plane.

The force moving the mass down the plane is gravity while Q is acting against it (hence the first equation)

(diagram showing the 8N weight being split up into the horizontal and vertical components)

If you find such an old post that was sorted years ago then it's best just to leave it and start a new one if you're stuck yourself.

And please read the Sticky post at the top of the forum about posting guidelines - we only post hints and suggestions, not full solutions :)

Have you found this website: https://www.drfrostmaths.com/sow.php...2017&term=Main

it's kind of intuitive. I like to think about it like vectors. I thought of z and w as position vectors. And so z-w is just the shortest distance between them. Idk if that makes sense to you. Maybe try drawing it. Not as and argand diagram. Just draw to random points z and w and work out what z-w represents