Lol cool. What do you think the GT will be? Above 60?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
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Lol cool. What do you think the GT will be? Above 60?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.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
ofcouse 63 or 62 in my opinion i hope so or itll be a BLol cool. What do you think the GT will be? Above 60?
my expectation was 74 because of that function mistake, now i have spotted 2 mark further mistake, in the perimeter of the circular measure.if that is right then if it wasnt for the Q7 i would have achieved 72 darn it :/
1.166whtz the answer for 10 (b)
What'd you do wrong in the perimeter?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 ;(
Me too did you write continuation?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
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 -bccosAWhat'd you do wrong in the perimeter?
Ow well things happen for a reason so ^.^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
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
Nope!!!!Me too did you write continuation?
In Sha Allah.hope the gt goes down inshallah ... just pray
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