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Zoes GCSE maths help

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Reply 40
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z_o_e
OP
11
Original post by notnek
Now you just need to combine all the like terms. You can deal with surds similar to how you deal with letters in algebra so what's this:

7x+6x−10x7x+6x-10x

?


I got 3 root 2 :smile:

Thank you so much

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Reply 41
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z_o_e
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Can someone please explain this

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Reply 42
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Notnek
21
Original post by z_o_e
Can someone please explain this

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You've missed part of the question.
Reply 43
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z_o_e
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Original post by notnek
You've missed part of the question.


Oh sorry find values of P and Q

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Reply 44
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Notnek
21
Original post by z_o_e
Oh sorry find values of P and Q

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Do you know how to find the turning point of a quadratic e.g. x2+4x−5x^2+4x-5 ?

Here you need to try and work backwards since you're given the turning point. Have a go at that and tell us if you get stuck. If you have no idea then that's fine, we can start you off.
Reply 45
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z_o_e
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11
Original post by notnek
Do you know how to find the turning point of a quadratic e.g. x2+4x−5x^2+4x-5 ?

Here you need to try and work backwards since you're given the turning point. Have a go at that and tell us if you get stuck. If you have no idea then that's fine, we can start you off.


I did this

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Reply 46
Avatar for Notnek
Notnek
21
Original post by z_o_e

Assuming the question was to find the turning points of y=x2−4x−5y=x^2-4x-5 then you completed the square correctly to get

(x−2)2−9(x-2)^2-9

But then this means the turning point is (2,-9) not (2, 9). You change the sign of the number in the brackets (-2 -> +2) and keep the sign of the other number (-9).


Okay now imagine the question was the reverse i.e. you are told the turning point is (2,-9) and were asked to find the quadratic equation. You would work backwards to get

y=(x−2)2−9y = (x-2)^2 - 9

And then expand this to finish. Can you use this method for your question?
Reply 47
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z_o_e
OP
11
Original post by notnek
Assuming the question was to find the turning points of y=x2−4x−5y=x^2-4x-5 then you completed the square correctly to get

(x−2)2−9(x-2)^2-9

But then this means the turning point is (2,-9) not (2, 9). You change the sign of the number in the brackets (-2 -> +2) and keep the sign of the other number (-9).


Okay now imagine the question was the reverse i.e. you are told the turning point is (2,-9) and were asked to find the quadratic equation. You would work backwards to get

y=(x−2)2−9y = (x-2)^2 - 9

And then expand this to finish. Can you use this method for your question?



I got x^2 -4x +4
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Reply 48
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Notnek
21
Original post by z_o_e
I got x^2 -4x +4
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Close but it's not quite right. Can you post your working?
Reply 49
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z_o_e
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Original post by notnek
Close but it's not quite right. Can you post your working?




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Reply 50
Avatar for Notnek
Notnek
21
Original post by z_o_e

I'm not sure what you've done here. I'll start you off:

The quadratic has a turning point at (2, 5). So if you work backwards the equation must be

y=(x−2)2+5y=(x-2)^2+5

If you expand then you'll get the answer.
Reply 51
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z_o_e
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11
Heya can someone help me on this plz
ImageUploadedByStudent Room1488649503.362190.jpg


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Original post by z_o_e
Heya can someone help me on this plz

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First consider the triangle ADC\text{ADC} (one of the faces of the tetrahedron) and use Pythagoras' Theorem to work out the length DC which is the hypotenuse. The two side lengths for this triangle are given to you.

Next consider the triangle ABD and again use Pythagoras to work out the length of the line DB as that is the hypotenuse.

Then consider yet another triangle ABC and work out the length BC using Pythagoras once more.

Finally, you should have all 3 sides of the triangle BDC. Apply the cosine rule and solve for angle BDC.

N.B. The first 3 triangles are all right-angled, while the last one isn't so you cannot apply any SOHCAHTOA on that one.
(edited 7 years ago)
Reply 53
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z_o_e
OP
11
Original post by RDKGames
First consider the triangle ADC\text{ADC} (one of the faces of the tetrahedron) and use Pythagoras' Theorem to work out the length DC which is the hypotenuse. The two side lengths for this triangle are given to you.

Next consider the triangle ABD and again use Pythagoras to work out the length of the line DB as that is the hypotenuse.

Then consider yet another triangle ABC and work out the length BC using Pythagoras once more.

Finally, you should have all 3 sides of the triangle BDC. Apply the cosine rule and solve for angle BDC.

N.B. The first 3 triangles are all right-angled, while the last one isn't so you cannot apply any SOHCAHTOA on that one.


Thank you I have got the answer:smile:

Any help on question B would be appreciated
ImageUploadedByStudent Room1488750292.473783.jpg


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Reply 54
Avatar for Notnek
Notnek
21
Original post by z_o_e
Thank you I have got the answer:smile:

Any help on question B would be appreciated
ImageUploadedByStudent Room1488750292.473783.jpg


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The next step is to factorise each of those groups.

So you can take a factor of 3x out of the first group and a factor of 14 out of the second group.
Original post by z_o_e
Thank you I have got the answer:smile:

Any help on question B would be appreciated



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The area of the lawn would be (the total area of the lawn + path) - (area of the path)

So, express area of the whole path in terms of xx firstly, then use the above to work out what xx is before finding the actual area of the lawn.

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