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The straight line L1 passes through the points with coordinates (4, 6) and (12, 2) The straight line L2 passes through the origin and has gradient -3. The lines L1 and L2 intersect at point P. Find the coordinates of P.

Find equation of line 1 in terms of x. eg y=mx+c - Using gradient and points. Equation of line 2 is just y=-3xSo -3x=mx+c of line 1Find xSub into one equation to find the y point.Done
RJ
Answered by Rishi J. • Maths tutor
10841 Views

A ball, dropped vertically, falls d metres in t seconds. d is directly proportional to the square of t. The ball drops 45 metres in the first 3 seconds. How many metres does the ball drop in the next 7 seconds?

d = kt 2 . 45 = k x 3 2 . 45 = k x 9 (/9). 5 = k.3+7 = 10 seconds total as we are investigating the NEXT 7 seconds. d = kt 2 . d = 5 x 10 2 . d = 5 x 100d = 500m in 10 seconds. As the ball drops 45m in the f...
EN
Answered by Emily N. • Maths tutor
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expand and simplify (x+1)(x-1)

first begin by expanding the brackets using the CLAW method (shown using whiteboard diagram)multiply the first of each bracket together, we get x 2 multiply the first term of the first bracket and the second...
AT
Answered by Adam T. • Maths tutor
11177 Views

root3 (root6 + root12) can be written as Aroot2 + B

root3 (root6 + root12) (root3 * root6 ) + (root3 +root12) = root18 + root36= (root9 + root2) + 6= 3root2 +6a= 3 b = 6
DM
Answered by Dorcus M. • Maths tutor
3962 Views

v^2 = u^2 + 2as u = 12 a = –3 s = 18 (a) Work out a value of v. (b) Make s the subject of v^2 = u^2 + 2as

v^2 = 12^2+2(-3)(18) = 36. Therefore by square rooting v = 6.To make s the subject first minus u^2 from both sides to have v^2 -u^2 = 2as then divide 2a from both sides to have (v^2-u^2)/2a = s
HN
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