Recent content by Simonio

Then I get: \(v(2v-5)\frac{80}{v}\ +v(2v-5)\frac{8}{3} = 160v\) = \(80(2v^2-5) + (2v^2-5)\frac{8}{3}\ = 160v\) = \(160v^2 - 400v + \frac{16v^2-40v}{3}\ = 160v\) Not sure whether I've lost the plot here!

Do you mean multiply by \(2v-5\)?

Not sure about the next bit: do I multiply throughout by \(2v-5\)? That would make: \((2v-5) \frac{80}{v} + (2v-5) \frac{8}{3} = 160\) Think I'm on the wrong track

OK-you've given me a good hint there! so: \(\frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}\)

Well I think I can say that \(t\)= \(\frac{80}{v}\) and with the slower speed \(t\) = \(\frac{80}{v-2.5}\) not sure of next step

I'm having problems getting going on the following question, any help appreciated: As part of his training an athlete usually runs 80 km at a steady speed of \(v\) km h. One day he decided to reduce his speed by 2.5 km h and his run takes him an extra 2h 40 mins. Derive the equation...

So: \(v^2 + 200v - 9600 =240v\) Then: \(v^2 - 40v -9600 = 0\) next, find factors of \(ac\) whose sum is \(c\) = 120, -80. \(v2 + 120v -80v - 9600\) Then: \(v(v-80) + 120(v-80)\) So either \(v+120 = 0\) or \(v-80 = 0\) Then \(v = -120\) or \(v = 80\) hmm... the answer in my book gives v...

Thanks. Could you show the working..so I know how you got there? Thanks.

Do we have to multiply throughout by \(v\)? Would that look like this: \(v(v-40)v\) \((\frac{240}{v}+1)\) Hmm...think I've gone wrong somewhere...

I'm having difficulty deriving a quadratic from the info in this question: A train usually covers a journey of 240 km at a steady speed of \(v \text{ kmh}^{-1}\). One day, due to adverse weather conditions, it reduces its speed by 40 \text{kmh}^{-1}\) and the journey takes an hour longer...

I'm looking at a question that refers to kilometres per hour and uses the abbreviation\(km h^-1\) It looks like a -1 index -could someone [point out to me what this means (I'm only just starting to get back into maths!). Thanks

Thanks (i'm trying to teach myself after many years of not doing maths so this help is invaluable): So \(4x(4x-33)-3(4x-33) Then: \((4x-3)(4x-33)\) So: \(x=\frac{3}{4}\) or \(x=\frac{33}{4}\) Is this now ok? I can see I had been sloppy before-I'm finding it's so easy to make small...

Thanks(Clapping)-so now I can say: \(16x^2-132x-12x + 99\) Then factorise in two groups: \(4x(4x-33)(-4x+33)\) then \((4x-3)(-4x-33)\) So: \(x = -\frac{-3}{4}\) or \(x = \frac{33}{4}\) Have I used the best method here?

I'm trying to factorize this: \(16x^2 -144x+99\) The only factors I can find of ac are: 132,12 which add up to -144 but only if both are negative and we need opposing signs to get the -144. I've used an online factor calculator and can't seem to find anything there! I'm obviously missing...

Thanks. Wow, I misread the meaning and thought that the 20x8 dimension was for the inner garden area -I find it's easy to get verbally confused! I'll have a go now based on your diagram -thanks a lot!