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Re: 回复: Locus. argh, didn't think of that

that one ^. it's locus this one is just (x+1)2 + (y-3)2 = 5

didn't we solve pretty much that exact question in the other thread? :confused:

thanks

clearly the solution for b) is wrong, since that says the gradient is 1 d) use perpendicular distance formula. We know the y coordinate is 0 |4x - 3y - 8|/[sqrt.a^2+b^2] = 6 2/3 4x+8/5 = 6 2/3 x = 6 1/3 (I think) you should be able to do the other questions

height of the water is proportional to the radius of the section filled at any point say radius/height = k then the formula for volume of a cone is 1/3 *pi*k^2*h^3 which can be expressed as bh^3, where b is a constant so we have: b*150^3 = 2b*h^3 h = 150/[2^(1/3)] =~119

ahhliss, your working is correct (same as mine at least) so I'm pretty sure there is something wrong with the question - maybe the values are incorrect?

lol you only need one line really 150/[2^(1/3)] =~ 119 since radius is proportional to height, h13 = 2h23

how can cos x be √(76)/76? look how small the adjacent is compared to the hypotenuse!!

:lol:

for your initial questions - if you know that SHM is the one dimensional motion of an object going around in a circle, you can quickly draw a diagram to find the angle when t=0 in the temperature questions, don't forget that a particular temperature will occur twice in the day

in the time it takes a car to travel 150m at 100km/h, a car travelling at 30km/h will have travelled 30/100 times that distance.. so 45

(1+x)^n = nCo x^0 + nC1 x^1 +...+ nCn x^n the total number of terms in the expansion before collecting like terms is 2^n - as you are picking either x or 1 (2 choices each time) n times Therefore nC0 + nC1 + nC2+...+ nCn = 2n. there's probably stuff missing for this to fully count as a proof though

as I said earlier, let C = A + B, then get rid of the A's [top]sin(C - B) + sinC + sin(C + B) = sinC(2cosB + 1) [bottom]cos(C-B) + cosC + cos(C + b) = cosC(2cosB + 1) top/bottom = sinC/cosC = tanC = tan(A+B)

start with (a+b)(c^2+?) now to get a^2(b+c) there must be an ab and ac in the (?), to multiply with the a (a+b)(c^2+ab+ac+?) When multiplied by b, this gives ab^2 + abc Now all that's left is cb^2 and (k-1)abc So add in a bc, to multiply with the b (a+b)(c^2+ab+ac+bc) Which also gives + abc when...

yeah just try and picture the point that moves at right angles to 2 points and it should jump out at you that "hey, it's a circle!"

wouldn't it just be (x+1)^2 + (y-2)^2 = 2 ?

The K method is used when you want to find the equation of the line that goes through the point of intersection of two lines, and another point. If the two lines are a1x + b1y + c1 = 0 and a2x + b2y + c2, and the point is (m,n), you write it as a1x + b1y + c1 + k(a2x + b2y + c2) = 0 then...

the point on line y=-1 will have coordinates (x, -1) by the distance formula.. 2 sqrt[(y- -1)^2 + (x-x)^2] = sqrt[(y-2)^2 + (x-3)^2] squaring both sides, and expanding gives 4y^2 + 8y + 4 = y^2 - 4y + 4 + x^2 - 6x + 9 3y^2 + 12y - x^2 + 6x - 9 = 0 or 3y(y+4) - (x-3)^2 = 0

oh I get it my method would be: convert your numbers into base 7 - ie number only have digits 0-6 the maximum value you can have is 50, which is 71 in base 7. now, a set will contain no pairs that sum to a number divisible 7 if none of the digits in the units of the two numbers sum to 7, or to...