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AFTER 24 hr .. P1 discussion here !

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thats cause i and a friend went and discussed our answers i got this whole question iguess wrong and i forgot one value of K rest was ok . well u can have an estimation
Lol cool. What do you think the GT will be? Above 60?
 
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thats cause i and a friend went and discussed our answers i got this whole question iguess wrong and i forgot one value of K rest was ok . well u can have an estimation
Same as me. Missed out on one value of k.
And got 1 question wrong.
But the one I got wrong was the one that mattered the most :/
Question 7... *sigh*
 
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if that is right then if it wasnt for the Q7 i would have achieved 72 darn it :/
my expectation was 74 because of that function mistake, now i have spotted 2 mark further mistake, in the perimeter of the circular measure.
disappointment ;(
 
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What was the Domain and range in function question ? I got
Y ≥ 1
-2.5 ≤ X and less than zero
 
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hey guys I cotinued my question on extra sheet does it cause any problem ....I mean i did half of question on answer sheet provided and half on extra sheet
please tell me:(
 
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hey guys I cotinued my question on extra sheet does it cause any problem ....I mean i did half of question on answer sheet provided and half on extra sheet
please tell me:(
Me too :( did you write continuation?
 
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What'd you do wrong in the perimeter?
it was an extremely foolish mistake, though we were given the square i forgot the fact and went on to calculate the lenght of BC by using cosine rule but I didnt used the correct formula. Instead of using a^2= b^2 + c^2 -2bccosA i used a^2= b^2 + c^2 -bccosA
I forgot the 2 in the formula and got the lenght of BC as 11.36 instead of the original lenght which was 10. hence my perimeter got incorrect :(
 
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it was an extremely foolish mistake, though we were given the square i forgot the fact and went on to calculate the lenght of BC by using cosine rule but I didnt used the correct formula. Instead of using a^2= b^2 + c^2 -2bccosA i used a^2= b^2 + c^2 -bccosA
I forgot the 2 in the formula and got the lenght of BC as 11.36 instead of the original lenght which was 10. hence my perimeter got incorrect :(
Ow well things happen for a reason so ^.^
 
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The question gives point R (-1.3) and line of 3y + 2x = 33 as the mirror. We are looking for the image of R, (let's name it as P)

Rearranging the equation of the mirror gives us y = -2x/3 + 11

Therefore gradient (m) of mirror is -2/3.

Since we understand that the line PR will be PERPENDICULAR to mirror (properties of reflection), therefore m of PR = 1.5.

Work out the equation of PR from the point (-1,3) and m of 1.5

y - 3 = 1.5 (x+1)

y = 1.5 x + 4.5

Another properties of mirror is that distance from object to mirror = distance from image to mirror. Therefore point of intersection of line PR and the mirror will be the MIDPOINT for line PR.

Working out simultaneous equation on PR and mirror,

y = 1.5 x + 4.5
y = -2x/3 + 11

1.5 x + 4.5 = -2x/3 + 11

9x + 27 = -4x + 66

13x = 39
x = 3

Subs x = 3 to any of the 2 equation above, we will get y = 9

Since we know R, and the midpoint, we can work out P.

(-1 + x)/2 = 3 for the x-coordinate and (3+y)/2 = 9 for the y-coordinate

Therefore x = 7 and y = 15


tanks bro ... but wy did tey give us this new concept""
and mashallah how did u attempt to do it... or was it a help from ur teacer
 
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