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That's amazing! I'm very glad

I agree! It's an unnecessarily hard question..

This is question 7, I hope it helps! th-cam.com/video/KVIjG9hcK-s/w-d-xo.html

I'm sorry, I'm not sure what you mean.. Is it the wrong link? The wrong question?

I'm very glad!

good luck, I just remembered the large data set 😭

Because 11 m is from A to the top (so when the particle is moving up). Then the particle moves down those 11 m and a bit further down. So you have 2x11 plus the extra bit. Does it makes sense?

@@betteryourmaths yes thank you

Thank you! My pleasure

Hi, yes, in the notes in the mark scheme says specifically to accept the n+1 working out, which means you would be using 128.5 instead of 128. For your second part of the question, you would find the mean between the 128th and 129th value when you have the actual values (raw data). When you have a grouped frequency table/histogram, you don't know the values, you assume they are evenly spread, so you are estimating instead of choosing a value. Which means you can use in the interpolation any value. In this question 128.5. I hope it makes sense

I just cancelled g and made s the subject. So you have first 6h/7=2s and when you divide by 2, you have 3h/7 = s

Well..you don't have the y-intercept..

@@betteryourmathsu can work it out tho?

Yes, absolutely. I just tend to use the formula I used in the question when I have gradient and one point and y=mx +c when I know gradient and y intercept. But it's me being silly😅 there is not a particular reason. You can find c substituting the point.

My pleasure!

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My pleasure! I'm glad it helped!

Yes it is!

Because while particle Q stops suddenly when it hits the ground, particle P keeps moving a bit up thanks to the velocity it already has. So it behaves as a particle projected up moving freely under gravity. Does it make sense?

I didn't use a CAST diagram, I usually use a circle of radius 1, I think it makes more sense and it's more difficult to forget. Do you know that method? But you can use CAST diagram if you prefer of even the graphs of sin, cos or tan to find other solutions. It's really up to you.

It's the same. I was unhelpful skipping steps, but when you solve b^2-4ac>0, you can find the roots of b^2-4ac = 0 and then decide if k must be greater or smaller than the roots.

Thank you!

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Is it? Thank you!😅

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You are "getting them into" the cubic root, so they have to go cubed. It's not a very mathematical explanation, it's difficult texting, I hope you understand what I mean!

@@betteryourmaths so if you put any number in a cube root, it will automatically become cubed? Same goes for squared root?

@@betteryourmaths Are there some general powers we need to know in the exams because I don't think I would be able to know that 6^4 is 1296 and put that in the equationat 5:09

Noooo...you have the calculator for all your papers!!😅

Yes, because a cubed root of any cubed number is the number. Yes, the same with square or any root

Thank you!

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Because you are working with half of the surfboard for the expression to be a function, so when you find the maximum, it's just half of the board width. To have the full width you need to multiply by 2.

@@betteryourmaths ohhhh, I read it again and it says Top half in bold, I got it thanks!