random maths as questions

do you mean sum?

its an arithmetic progression, with a=1 and d=1 and n=300, => sum = n/2 (first + last) = 150*(1+300) = 45150

yeah AP sn= n/2 (2a (n-1)d)

but how do u no theat d=1, dnt get that.

btw, you seem very cleva in maths.

becuase it is 1+2+3+4+5+...+298+299+300. the difference between each term is 1

elpew, if it asks you to evaluate

300
e (sigma) (r2 (squared) + r)
r=1

Youve just evaluated SUM {r=1,300} r , so you need to add on SUM {r=1,300} r^2. This sum you can find in the formula book, and is n/6(n+1)(2n+1)

so you basically join up what above sigma with what below, put them in brackets, multiply by r^2 and check it up in the formaul book, and itll give you the answer??

how bout if it asks...

the points A, B and C have coordinates (1,7) (5,5) (7,9)

a) show that AB and BC are perpendicular.
b) find the equation of BC
c) the equation of the line AC is 3y= x + 20 and M is the midpoint of AB
i) find an equaiton of the line through M parellel to AC.
ii) this line intersects BC at the point T. find the coordinates of T?

thanks to any1 who answeres and explians their answers.

a) The gradient of a (segment of a) line is given by (change in y)/(change in x) so the gradient of AB is (7-5)/(1-5) = -1/2, and the gradient of BC is (5-9)/(5-7) = 2. Two lines are perpendicular if the gradient of one is m, and the gradient of the other is -1/m, so AB and BC are perpendicular.

b) The general equation of a line with gradient m and going through the point (a,b) is (y-b) = m(x-a) Youve just found 'm' and both B and C lie on the line

c) The midpoint of two points (a,b) and (c,d) is ( (a+c)/2 , (b+d)/2). Use this to find M. The gradient of AC is 1/3. Therefore the gradient of the line you want is 1/3.

When two lines intersect, their x coordinates are the same, and their y coordinates are the same. Make simultaneous equations and solve to find x and y.

Post back if you need more help

a) The gradient of a (segment of a) line is given by (change in y)/(change in x) so the gradient of AB is (7-5)/(1-5) = -1/2, and the gradient of BC is (5-9)/(5-7) = 2. Two lines are perpendicular if the gradient of one is m, and the gradient of the other is -1/m, so AB and BC are perpendicular.

b) The general equation of a line with gradient m and going through the point (a,b) is (y-b) = m(x-a) Youve just found 'm' and both B and C lie on the line

c) The midpoint of two points (a,b) and (c,d) is ( (a+c)/2 , (b+d)/2). Use this to find M. The gradient of AC is 1/3. Therefore the gradient of the line you want is 1/3.

When two lines intersect, their x coordinates are the same, and their y coordinates are the same. Make simultaneous equations and solve to find x and y.

Post back if you need more help

btw, i didnt get the bit when you said 'Two lines are perpendicular if the gradient of one is m' how do i know what m is. do i have to go through the process of working out m= (y2-y1/x2-x1)?

if two lines A and B are perpendicular,then m(A)*m(B)=-1 i.e. m(A)=-1/m(B).

yeah, thx alot for that. you seem like a very bright person regarding, appreciate it alot pal

Lol, eplaw building up a fanbase

He's the best Physicist in Oxford.

he goes oxford??? wow? that guy must be real clever. how do you know?