Urgent Help need on Maths GCSE Questions Watch

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Melanie47
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#1
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I scanned in the question - there are others but I'll put the links up to them later.
Click here for the question.

Thanks for any help given. You can pm me if it is more convenient. I will be back on after 9 o'clock.
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J.F.N
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p=3^8, so p^0.5=3^4

q^1/3 = 2^3 . 5^-2

q^-1= 2^-9 . 5^6
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Melanie47
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(Original post by J.F.N)
p=3^8, so p^0.5=3^4

q^1/3 = 2^3 . 5^-2

q^-1= 2^-9 . 5^6
Thanks so very much!
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Melanie47
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Okay, here are the rest of the questions:

Question 1

Question 2

Question 3

Question 4

Question 5
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Melanie47
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Can anyone help me, please? I know that there are a lot of questions, but I would be really grateful for any help
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manps
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q2

using SOHCAHTOA

cos23 = 6/hyp
hyp = 6/cos23

hyp = 6.52 3s.f
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manps
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q3

me thinks

total up all year groups = 1320

year 12 = 210 * (50/1320) = 8 students (to nearest student)
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manps
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q4

(i) 2x angles in same segment
(ii) ODA = OAD which is 2x+ x = 3x
so AOD is 180 - 6x (angles in a triangle add up to 180 degrees)
(iii) (180 - 6x)/2 angles at circumference are half that at the centre
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Melanie47
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Wow, thanks so much
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manps
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q5
BC = BP
AB = QB
QC = AP

SSS congruency

someone verify this
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Melanie47
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maps I cannot thank you enough for your help
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J.F.N
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(Original post by manps)
q5
BC = BP
AB = QB
QC = AP

SSS congruency

someone verify this
thats fine, just explain why QC=AP since its not immediately obvious.
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Melanie47
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(Original post by J.F.N)
thats fine, just explain why QC=AP since its not immediately obvious.
Okay.
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manps
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it's manps :rolleyes:

also q1 i 4got about

(a) smaller circle area = pi r²
larger circle area = pi R²
shaded region = pi R² - pi r²
since shaded region is equal to smaller circle, we can say,
pi r² = pi R² - pi r²
2pi r² = pi R²
2r² = R²
R = root2 r

(b)

think of a right angled triangle
Call midpoint of AB, Z
BZO = 90degrees so you know what the triangle looks like
hypotenuse = root2 r
OZ = r and say ZB = x
using pythagoras,
(root2 r)² = r² + x²
2r² = r² + x²
r² = x²
r = x
since ZB is half the line, AB = 2*x which is 2x and we know x = r so 2x = 2r QED
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manps
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(Original post by Melanie47)
Okay.
not quite sure how to explain it but if triangle RQC is placed on top of BAP then QB=AB and BC=BP which therefore must mean that QC=AP otherwise the other 2 sides cant match

any1 who would like to comment further is welcome to
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SsEe
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q5. I'll explain everything in my method

AngleQBA=60 and AngleCBP=60
in TriangleABP you have AngleABP = AngleABC + AngleCBP = AngleABC + 60
in TriangleQBC you have AngleQBC = AngleABC + AngleQBA = AngleABC + 60

AngleQBC = AngleABP

Also QB=AB and BC=BP. Picture one of the triangles rotating onto the other. The sides will match and the angle will match so it's a SAS thingy.
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Melanie47
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#17
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Thanks for all the help given, but just to let others know, I have handed in the paper now
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