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The time has to be between 0 and 7, otherwise sqrt(49-t^2) will not be defined and time can not be negative. At t=0 x=0 and t=7 x=0. And x is positive as both t and sqrt(49-t^2) is positive. So if we sketch x versus t it will be from zero to a max and then back to zero. To find that max...

For part c) of the first question: The sale of the 1st month of graphics scientific calculators is 2000(10% of 20000). It goes up by 150 so d=150. T(n)=a+(n-1)xd We want T(n) to be 20000 so 20000=2000+(n-1)x150 will give n=121. For question 10 (or the second question you posted) : Mark a point...

I put the answer on my website. Please scroll all the way down the page. www.cambridgecoaching.com.au Reza Bokat, www.cambridgecoaching.com.au

Yes, but we are talking about 2 unit course, not 3 unit or 4 unit. Each course requires its own approach. Reza Bokat, www.cambridgecoaching.com.au

Since the chance of catching a legal length is 4/5, the chance of not catching one is 1/5. Since we talk about 3 catches then: 4/5x1/5x1/5=4/125 is the chance of 1st catch only being a legal length 1/5x4/5x1/5=4/125 is the chance of 2nd catch only being a legal length 1/5x1/5x4/5=4/125 is the...

secx=0 has no solution. (If you look at the graph of secx it is always greater or equal to one or less than or equal to -1, in other words secx cannot take any values between -1 and 1, so it cannot be zero). Also 2sec^2x-1=0 has no solution. Similarly sec^2 x will be always greater than one so...

a) The diameter will contract in the direction of motion but unchanged perpendicular to the motion. In the direction of motion the diameter will be 120000xsqrt(1-(0.25c/c)^2)=116190.5 km. b) The length of the Klington vessel as seen from earth will be contracted, so multiply its stationary...

For mission control, it is exactly like you have a car moving at 100 km/hr, from Sydney to Canberra ( 300 km away) . The time for this trip is time=distance/speed t=300/100 or 3 hours. In the same way t=4.5/0.9 which gives 5 years for mission control. ( If distance is in light year and speed is...

t0=2.2x10^-6 t=1.6x10^-5 v=? t=t0/sqrt(1-(v/c)^2) After substiution: 1.6x10^-5=2.2x10^-6/sqrt(1-(v/c)^2) 2.2x10^-6/1.6x10^-5=sqrt(1-(v/c)^2) 1-(v/c)^2=0.01890625 (v/c)^2=0.98109375 v/c=0.99 v=0.99c but c=3x10^8 m/s so v=2.97 x10^8 m/s I hope it is clear. Reza Bokat Cambridge...

The solution for 6d is as follows. By odd multiples of 100 they mean: 1x100, 3x100, 5x100 etc. So the terms are 100,300,500,....,(2n-1)x100 The term no. n being (2n-1)x100 ,a=100 and d=200 Using the sum formula s=n/2(2a+(n-1)d) we get: s=n/2(2x100+(n-1)200) which gives s=100 x n^2 I hope it...

Find the gradient of PS (S is the focus.). Find the gradient of QS. Equate the two gradients to show that pq=-1. Use pq=-1 to show that the x coordinates of U and R are equal. Hope it helps. Reza Bokat, Cambridge Coaching www.cambridgecoaching.com.au

a) Sub zero for x in f(x) to get f(0)=-1 Sub one for x in f(x) to get f(1)=3 As f(x) changes from negative to positive between 0 and 1 so there is a root for f(x) between those values of x. b)Differentiate f(x) to get f'(x)=6x^2+2. This is always positive for all values of x which means f(x) is...

e^x is the derivative of itself. In other words when you differentiate e^x, it does not change.

You need to use the quotient rule which says if y=u/v then y'=(u'v-uv')/v^2 In this case u=x^2 so u'=2x and v=e^x so v'=e^x So y'=(2x.e^x-x^2.e^x)/(e^x)^2 After factorising for e^x at the denominator and numerator and eliminating it you get: y'=(2x-x^2)/e^x. We need y' when x=0 so sub 0 for x...

Write A as below: A=500-10x-4000x^-1 Differentiate: dA/dx=-10+4000x^-2 Equate this to zero. -10+4000x^-2=0 and solve to get: x=20 We do not take x=-20 as x is length and cannot be negative. Find the second derivative to check to see if this value of x gives the max value: A''=-8000x^-3 Now...