IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: al noor on November 08, 2009, 03:07:12 pm
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Can any1 solve these problems for me as i want to check if my answrs are right ;)
Thanyou in advance :D
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Okay, will try to help solve it for you. :) Give me a while.
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bridge question a)50
b)250(500-250)800+60
These numbers seem incredible. For the next part I get a ridiculous percentage error.
Are you sure these questions are correct?
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It says y = x(500-x)800 + 50
should it say y = x(500-x)/800 + 50
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b)i)30
ii)400
iii)10 and 50
iv) presumably x-60 so loss =-60^2+50+50-500=-500
vi)-x(x-30)^2+400
max profit of 400 at x=30
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bridge question a)50
b)250(500-250)800+60
These numbers seem incredible. For the next part I get a ridiculous percentage error.
Are you sure these questions are correct?
Well thats the worksheet that the teacher gave us
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b)i)30
ii)400
iii)10 and 50
iv) presumably x-60 so loss =-60^2+50+50-500=-500
vi)-x(x-30)^2+400
max profit of 400 at x=30
Thnkz aloott astar
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Here you go! For the first page :)
Dear astarmathsandphysics,
Would you mind helping me check it as well, sir? xD Just in case I make mistakes. By the way, the equation is fractional :)
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Dear al noor,
For the 2nd question part (a), once again, your graph is a parabolic curve with the symmetry at x=30. Therefore, that's where your maximum point will lie which is (30,400). It should begin at -500 and end there as well therefore, the starting point of the graph would be (0,-500) and the end point would be (60,-500)
EDIT : For the 2nd question part (c), the expression should be (x-30)^2 -400. There shouldn't be a -x outside. The connection is as stated by astar. :) I'm trying to figure out the part where they ask for the loss when toys are given away though.
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It should say y = x(500-x)/800 + 50
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It should say y = x(500-x)/800 + 50
Yep! I opened it with Microsoft 2007 and it shows a fractional equation :)
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c)%error =(121-115.625)/121*100
dii) steepest when dy/dx greatest or least y=5x/8-x^2/800+50 so dy/dx=5/8-x/400 so greatest when x=0 or 500
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Here you go! For the first page :)
Dear astarmathsandphysics,
Would you mind helping me check it as well, sir? xD Just in case I make mistakes. By the way, the equation is fractional :)
Thankyoouu soo muccchh :)
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Thankyoouu soo muccchh :)
You're most welcome. Hope it helps :)
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Thankyoouu soo muccchh :)
can u help me with these also?
It will be a great pleasure ;)
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Here you go! The other document will be posted once I'm done :)
EDIT : 7.docx is completed!
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I hope this doesn't show spam but.. al noor! I'm done! Refer to previous reply :)
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Here you go! The other document will be posted once I'm done :)
EDIT : 7.docx is completed!
Thankyouuu soo mucchhh ;)
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No prob :) hope it helps!
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No prob :) hope it helps!
It helped me soo mucch.:D:D:D:D
Thankyou and thankyou astar
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I have a question
Show that the curves y=2x^2 + 5x , y= x^2 + 4x + 12, y=3x^2 + 4x - 6 have one point in common and find its coordinate its from the book P1 Mathematics ( AS LEVEL CAMBRIDGE)
thanks
i need an urgent solution pleasee..
cheers:D
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1)y=2x^2 + 5x
2)7=x^2 + 4x + 12
3)y=3x^2 + 4x - 6
1)=2) 2x^2 + 5x =x^2 + 4x - 12
x^2+x-12=0=(x+4)(x-3) so x=-4,3
1)=3)2x^2 + 5x =3x^2 + 4x - 6
so x^2-x-6= 0 so (x+2)(x-3)=0 so x=-2,3
2)=3)x^2 + 4x +12=3x^2+4x-6 so2x^2-18=0 so x^2=9 so x=3, -3
ans is x=3
y=x^2+4x+12=9+12+12=33
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hey, thanks alot it really helped.
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I have another question
The line y = 6x + 1 meets the curve y=x^2 + 2x + 3 at two points. Answers are in surds..
source: Q22 P1 AS LEVEL maths cie
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6x + 1=x^2 + 2x + 3
x^2-4x+2=0
x=(-b+_sqrt(b^2-4ac))/2a=(4+-sqrt(16-8))/2=2+-sqrt(2)