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a mole of a gas at STP has a volume of 22.71L. molar mass/mass = molar volume/volume this gives you a molar mass of 72.99 for Chlorine gas, however as Cl gas is diatomic (Cl2) you divide by 2, = 36.49 atomic mass. Reasons for the different atomic mass (Cl's atomic mass is 35.45) that I can...

that simplifies to (x + 2)^2 + y^2 = 4 just half the radius of the old circle to find the midpoints, and change x and y in accord with the new centre this one? 0 < x < 3 0 < y < 3 the y bit means that the since the ladder will only be above the ground, y >= 0. I'm not sure how you would know...

Re: 回复: Re: Trigonometric Equations needed same thing isn't it, since pascal's triangle is just a way of expressing combinations? for lower values it would be easier to just have rows of the triangle memorised

done moving about and world communicates, now on electricity - half way through maybe?

P is a collinear point, not a locus btw It either divides the line AB internally or externally in the ratio of 2 to 5, if internally, then point P is made up of 2/7 of point B and 5/7 of point A (it approaches B from A and gets 2/7 of the way there, therefore with 5/7 of the way to go and thus...

well you could express that mathematically, ( 15Cr*2^16-r * 3^r-1 )/8 = (15C(15-r)*2^r-1 * 3^16-r)/27 2^16-r * 3^r-1 * 3^3 = 2^r-1 * 3^16-r * 2^3 16 - r = r-1 + 3 2r = 14 r = 7

you know that the rth term from beginning and end will have the same combiantions part of their coefficient and you can see that 8 = 2^3 and 27 = 3^3, so rth term from the end must be 3 places higher than the rth term, and since they each have the same number of places, x, on the other side of...

sin(7x + 4x)sin(7x - 4x) = (sin7xcos4x + sin4xcos7x)(sin7xcos4x - sin4xcos7x) = sin^2 7x cos^2 4x - sin^2 4x cos^2 7x = sin^2 7x (1 - sin^2 4x) - sin^2 4x (1 - sin^2 7x) = sin^2 7x - sin^2 4x sin^2 7x - sin^2 4x + sin^2 4x sin^2 7x = sin^2 7x - sin^2 4x it's the line before the bolded line that...

you

oh how I want to believe that this guy's just trolling :(

well I guess it depends whether the subjects you're doing are the type that you can majorly screw up on. you can be the judge of that. I'm personally gonna dropdown to 10 units - gives you a lot more time to work on the other units anyway you seem to have made a hell of a lot of 'stupid errors'...

what type of mistakes did you make in your maths tests, and why did you make them? it shouldn't be too hard to catch up with bio if you're pretty good at self-teaching. are you sure you can change subjects at this point in the year though? you'll be able to drop 2 units at the end of year 11...

it says biggest rectangle, not any quadrilateral so you shouldn't need to hesitate before giving that answer * let the sides of the rectangle be x and y let d = diagonal of rectangle = diameter of circle (therefore a constant) /would proof be needed that the diagonal of the rectangle is...

well, even if it was a prove question you could have made things easier by proving that sin@ = 2t/(1 + t^2). though wouldn't you be able to use the t identities in a proof?

the question is 2tan15/(1+tan^2 15) , and sin@ = 2t/(1+t^2), where t = tan @/2 so let tan @/2 = tan15, @ = 30 therefore 2tan15/(1+tan^2 15) = sin30 = 1/2 this is definitely an easy question since its just sub'ing into the t formula for sin@

from the t formulas, you should know that sin@ = 2t/(1 + t^2)

it probably would have taken less time to prove it than to write that shouldn't you be able to easily figure out the relationship between the arc length and r? since you know the arc is @/2pi * 2pi r, or don't you learn radians in year 11? what's question 24?

yeah I understand the equating coefficients thing, that was what I first thought to do. what threw me off was the question - it states the polynomials are equal for 3 values of x. Is this just another wasy of saying that the polynomials are equal for all x? since afaik you can't have 2 different...

that's what my working is too. I don't see anything wrong with it

putting the polynomial as k(x-1)(x+3)(x+a) when x = 2 5k(a + 2) = -15 k = -3/(a+2) when x = -1 -4k(a - 1) = 36 -4*-3(a-1)/(a+2) = 36 a - 1 = 3(a + 2) -2a = 7 a = -7/2 polynomial is therefore 2(x-1)(x+3)(x-7/2) removing the fraction: (x-1)(x+3)(2x-7) how do you know to sub in those values...