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    find the intergers from 1 to 300 inclusive?

    can any one help me do this please, il be very greatful, thanyou.
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    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
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    (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.
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    (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
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    (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
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    elpew, if it asks you to evaluate

    300
    e (sigma) (r2 (squared) + r)
    r=1
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    (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)
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    (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??
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    (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.
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    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.
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    (Original post by maximusmak)
    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
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    (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.
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    (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)?
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    (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).
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    (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
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    (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.
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    Lol, eplaw building up a fanbase
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    (Original post by mik1a)
    Lol, eplaw building up a fanbase
    I wasn't joking.
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    (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?
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    (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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