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# random maths as questions watch

1. find the intergers from 1 to 300 inclusive?

can any one help me do this please, il be very greatful, thanyou.
2. 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
3. (Original post by elpaw)
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.
4. (Original post by maximusmak)
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
5. (Original post by elpaw)
becuase it is 1+2+3+4+5+...+298+299+300. the difference between each term is 1
yeah, thx alot for that. you seem like a very bright person regarding, appreciate it alot pal
6. elpew, if it asks you to evaluate

300
e (sigma) (r2 (squared) + r)
r=1
7. (Original post by maximusmak)
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)
8. (Original post by It'sPhil...)
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??
9. (Original post by maximusmak)
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??
I don't understand what you're saying, but what It'sPhil meant was to find the sum of (r^2 + r), you find the sum of r^2 (to do that, use the formula in the formula book) and add it to what you've already worked out for the sum of r.
10. 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?

11. (Original post by maximusmak)

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?

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
12. (Original post by It'sPhil...)
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
very clever, i wish i was as bright as you mate, ill be flyin through the exam. thanks alot, amte, i really appreciate it. yeah sure, when i get stck on the paper ill ask you.
13. (Original post by It'sPhil...)
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)?
14. (Original post by maximusmak)
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).
15. (Original post by IntegralAnomaly)
if two lines A and B are perpendicular,then m(A)*m(B)=-1 i.e. m(A)=-1/m(B).
thx
16. (Original post by maximusmak)
yeah, thx alot for that. you seem like a very bright person regarding, appreciate it alot pal
He's the best Physicist in Oxford.
17. Lol, eplaw building up a fanbase
18. (Original post by mik1a)
Lol, eplaw building up a fanbase
I wasn't joking.
19. (Original post by Renaissance)
He's the best Physicist in Oxford.
he goes oxford??? wow? that guy must be real clever. how do you know?
20. (Original post by maximusmak)
he goes oxford??? wow? that guy must be real clever. how do you know?
Because he says he goes to oxford, his profile says he's doing a 4 year masters physics degree at oxford, his name is registered on the oxford computing website and he's a genius.

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