Urgent Help need on Maths GCSE QuestionsWatch

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Thread starter 14 years ago
#1
I scanned in the question - there are others but I'll put the links up to them later.

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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14 years ago
#2
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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Thread starter 14 years ago
#3
(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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Thread starter 14 years ago
#4
Okay, here are the rest of the questions:

Question 1

Question 2

Question 3

Question 4

Question 5
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Thread starter 14 years ago
#5
Can anyone help me, please? I know that there are a lot of questions, but I would be really grateful for any help 0
14 years ago
#6
q2

using SOHCAHTOA

cos23 = 6/hyp
hyp = 6/cos23

hyp = 6.52 3s.f
0
14 years ago
#7
q3

me thinks

total up all year groups = 1320

year 12 = 210 * (50/1320) = 8 students (to nearest student)
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14 years ago
#8
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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Thread starter 14 years ago
#9
Wow, thanks so much 0
14 years ago
#10
q5
BC = BP
AB = QB
QC = AP

SSS congruency

someone verify this
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Thread starter 14 years ago
#11
maps I cannot thank you enough for your help 0
14 years ago
#12
(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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Thread starter 14 years ago
#13
(Original post by J.F.N)
thats fine, just explain why QC=AP since its not immediately obvious.
Okay.
0
14 years ago
#14
it's manps 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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14 years ago
#15
(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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14 years ago
#16
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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Thread starter 14 years ago
#17
Thanks for all the help given, but just to let others know, I have handed in the paper now 0
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