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Let u = sinx Solve the quadratic in u, then replace u with sinx

Firstly it should be 1/e^x Secondly, after switching x and y, let u = e^y, and then solve the quadratic in u.

Just a practice essay done for homework.

For question 3, the two different equations are one and the same, as it is a common tangent. Equation 1: xsec@/a - ytan@/b = 1 Equation 2: x + t^2y = 2ct Dividing by 2ct, x/2ct + yt/2c = 1 This is the important step: equate coefficients of x and y, as the equations are exactly the same...

I will get tired of saying this eventually, but you cannot use those formulas from physics, and I doubt they will give them to you in questions. Consider only vertical motion: a = -g, integrate with respect to time and go from there.

differentiate parametrically to get the gradient: dy/dx = -bcos@/asin@ then use pt gradient formula: y - bsin@ = -bcos@/asin@ (x - acos@) asin@y - absin^2@ = -bcos@x + abcos^2@ yasin@ + xbcos@ = ab(cos^2@ + sin^2@) yasin@ + xbcos@ = ab ii. sub (ae,0) into the equation: bcos@ae = ab...

Let a = tan(inverse)3/5 and b = Tan(inverse)1/5 Then take tan of LHS of equation above: tan(a=b) = (tana + tanb)/1 - tanatanb Remember tantan(inverse)3/5 = 3/5, and similarly tanTan(inverse)1/5 = 1/5 so ... = 3/5 + 1/5 / (1 - 3/5.1.5) = 1 = RHS

1. domain: -1<=2x<=1 -1/2<=x,+1/2 range: its three time the range of the normal inverse cos function, ie 0<=y<=3pi 2. Sub in the two point and find the y values, then use rise/run to find gradient

i the tangent does have equation bxcos@ + aysin@ = ab differentiate: dy/dx = -bcos@/asin@ then use point gradient formula, multiply out and simply ii. sub in (ae,0): so bcos@ae = ab cos@ = 1/e So x-coordinate of P is a/e, so P lies on the directrix of the hyperbola iii) We know cos@ =...

for b, c, keep using newtons until successive apps agree to one decimal place, although this isn't strictly an error-proof method show d graphically

To find the pt of intersection you have to solve the equation 2 cos (x/2) = x Therefore let f(x) = 2 cos (x/2) - x, and bash away

That list looks like enough, but if you want more: -An intro with a brief discussion of the theory behind the experiment -Possible errors in the experiment -How the experiment can be improved

man i feel sorry for you guys. i dont know how our school got around it, but we always did stanard exams. good luck to all.

haha isn't my essay enough?

I don't know if this is the best method, but: Let S be the midpt of LM. Construct RS. Prove SR is parallel to MN (easily done considering triangles LSR and LMN are similar) Now we know angle RSM = QMP Also RS/SM = QM/MP (prove before hand that RS = 1/2MN from the same similar...

stop confusing people with your sophisticated legalese mr lubo

for a circle you'd need a + instead of the - its a rectangular hyperbola.

try x^2 - y^2 = 9, or something like that. i guess its a little advanced for prelim, but if you do 4u you'll need to use a method like trebla's later on.

That's the best method for vertical/horizontal asymptotes. But what about for oblique asymptotes? Trebla's method works for those.

if you need further help on 10: you will get two expressions, Vsin@ and Vcos@. Divide and therefore calculate tan@. Then sub @ back in for V.