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Mathematics: Post your doubts here!

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Part b Solution
360 x 2/3 = n^2 +3n +2
240= n^2 + 3n +2
so
n^2 + 3n + 2 = 240
n^2 + 3n + 2 -240= 0
n^2 +3n -238 = 0
n^2 +17n -14n - 238 =0
n(n +17)-14(n+17) = 0
(n+17)(n-14)=0
Since n cant be negative so n-14= 0 so n is 14
xcus eme is this te one i asked for
 
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Just solve normally..... the fg(x)

(2x+7)^2 + 2x+7 - 3
2x(2x+7) +7(2x+7) +2x+7 - 3
4x^2 + 14x + 14x + 49+ 2x + 4
4x^2 + 30x + 53

compare this with the original equation
4x^2 + 30x + 53 = px^2 + qx + r
in place of p = 4
in place of q= 30
in place of r= 53
oh thnx
acctually i didnt substitute d second x
 
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correct to 1 dp means the error cud be 0.5 ie( 1 divided by 2)
it can be plus or minus 0.5
so
for lower u minus 0.5 and for upper add 0.5
use the formula fr peri and find out
if u dont get this tell me !
for eg a question says a tyre has a daiameter of 12cm to the nearest cm. calculate upper and lower bound?
the 1st thing u do is divide the unit u have by 2
(1/2)=0.5
upper bound =12+0.5=12.5
lower bound=12-0.5=11.5
the thing that i don't get is why do we divide it by 2???
 
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GUYS NEED HELLPPP........

I'm stuck on this question, from O580/43/O/N/13 (October November 2013, 43, IGCSE)

It's Question 9ci and 9cii

c) The total number of lines in the first n diagram is 1/2n^3 + pn^2 + qn

i) When n=1, show that p+q=8.5


ii) By choosing another value of n and using the equation in part (c)(i), find the values of p and q.

Need help ASAP.

Link to the paper: http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
 
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GUYS NEED HELLPPP........

I'm stuck on this question, from O580/43/O/N/13 (October November 2013, 43, IGCSE)

It's Question 9ci and 9cii

c) The total number of lines in the first n diagram is 1/2n^3 + pn^2 + qn

i) When n=1, show that p+q=8.5


ii) By choosing another value of n and using the equation in part (c)(i), find the values of p and q.

Need help ASAP.

Link to the paper: http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
First one is easy! substiture 1 in the formula and the formula = 9 as the n is 1
for second one use simultaneous equation
 
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perimeter of rectangle=2L+2W
2(23.7)+2(10.9)=69.2
hence it is correct t 1 d.p (1/2)=0.5
L.b=69.2-0.5=68.7
U.b=69.2+0.5=69.7
:)
umm.. i checked the marking scheme and your answer is wrong i also did the working it's supposed to be:-
U.B 23.7+0.5 =23.75 10.9+0.5=10.95
23.75+23.75+10.95+10.95=69.4m
L.B 23.7-0.5=23.65 10.9+0.5+10.85
23.65+23.65+10.85+10.85=69m
 
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umm.. i checked the marking scheme and your answer is wrong i also did the working it's supposed to be:-
U.B 23.7+0.5 =23.75 10.9+0.5=10.95
23.75+23.75+10.95+10.95=69.4m
L.B 23.7-0.5=23.65 10.9+0.5+10.85
23.65+23.65+10.85+10.85=69m
SOORRRY i found the perimeter then did the U.b and L.b.
but actually we're supposed to find the U and L bound and then find the perimeter.
i forgot that i repeat my apology. Sorry.:(
 
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First one is easy! substiture 1 in the formula and the formula = 9 as the n is 1
for second one use simultaneous equation

I got the first one after some thinking :p

But the second part (9 cii) I can't seem to figure it out. It's not simultaneous equations cuz I looked at the mark scheme and atleast understood that much. Could you try solving it? I'd be ever so grateful! It's for 5 whole marks! :/
 
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