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Solve 10 -3 (2x-1) = 3x + 1
The answer is 1/13..but mine is coming as 4/3..
The answer is 1/13..but mine is coming as 4/3..
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Sure, Askcan anyone help me with Loci?
Follow the steps. : http://www.mathopenref.com/constbisectangle.htmlhow to draw an angle bisector?
ThanksOh that, i did that just a few hours back, dont go for the marking scheme... Seems like a mistake.. Even i got 4/3 and my teacher also told its correct!!
yes me too got 4/3Solve 10 -3 (2x-1) = 3x + 1
The answer is 1/13..but mine is coming as 4/3..
Nov 11yes me too got 4/3
which year ??
(i) 3y+ 4x= kI got another question :
A is the point (0, 2).
(i) The equation of the line AB may be written 3y + 4x = k.
Find the value of k.
Answer is 6.
(ii) Find the coordinates of the midpoint of AB.
For this..how do you find the coordinates of B from the equation?
No i get that..the second part..(i) 3y+ 4x= k
so place x= 0 and y= 2
than we get k=6
I got the part about the gradient..but what after it?..find gradient which is -4/3 and put them equal to (y-20/(x-0) to get x,y and then apply formula for midpoint
x - (20/100)x = 60The price of a camera in the sale was $60.
Calculate its marked price if the goods were reduced by 20%?
it is 4/3...Solve 10 -3 (2x-1) = 3x + 1
The answer is 1/13..but mine is coming as 4/3..
Do check the examiner reports...it has the correct answer 4/3 .Solve 10 -3 (2x-1) = 3x + 1
The answer is 1/13..but mine is coming as 4/3..
I got another question :
A is the point (0, 2).
(i) The equation of the line AB may be written 3y + 4x = k.
Find the value of k.
Answer is 6.
(ii) Find the coordinates of the midpoint of AB.
For this..how do you find the coordinates of B from the equation?
3y=-4x+6Well you wind mid point of any line by , y2+y1/2 x2+x1/2
you can find b as you now know what the k is , You know that the graidet is -4/3 .Just replace one random number of X and youll get your Y and hence you can use the formula to answer it . If you want ill answer it on the forums for you .
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