Funnily enough it was your answer of 45 that made me think properly about this one. It's got loads of irrationals from the roots, but somehow they must all cancel out and leave a simple number. So I started to think seriously about how that might happen.
If you times each term by a fraction thus
then the denominators simplify to just 2. Thus the roots are moved to the numerators and with a minus sign now instead of a plus. It's a bit like rationalising a surd.
Now put all over the common denominator of 2, and you get a load of cancelling like this
The only terms that don't cancel are -root(100) at the start and + root(10000) at the end. As both have integer roots that does it.
1) Use the difference of two squares thus:
Do that for every term, cancel all but the first and last, and it comes out (as 45, thanks Phro )
4)Let that expression = N
Using the quadratic formula you want b^2 - 4ac to be a perfect square.
So you just need to find the biggest t (under 10000) that gives such a perfect square.
Wow! I see your algebra has come on a lot.
1) and 4) are beyond my powers but others will hopefully jump in on those and give you (and me) an idea what to do.
2) I don't have a super number cruncher for this, so I'd have to slog through elimination. Do-able but tedious.
3) Arhh! One I can contribute to. If you expand (p+q+r)^2 you'll find you can re-arrange this so that pq+qr+rp is the subject and the other side of your equation can be substituted with numbers. So it's easy from there.
Might have an idea for 2). I'll be back if it works.
LATER EDIT: Yes it does.
Add all five equations and simplify to get the value of a+b+c+d+e
Now subtract the first from the second, substitute in for a+b+c+d+e and you'll get the value of e.
Do similar tricks with subtractions of other pairs of equations to get other letters.
When I was taught this it was called Newton's experimental law. If two balls collide with velocities u1 and u2 before the impact and v1 and v2 afterwards, then the coefficient of restitution is given by
and the law of conservation of momentum is
In real life 0 < e < 1. But the kinetic theory of gases assumes that for molecules e = 1.
For the impact of the ball with the ground, e must be 1 for otherwise the ball will not climb back to its original height. As we are not told the value of e for the impact with the block, I am assuming it, also, must be 1. There's no way to do the problem without knowing e, because it affects the velocity that the ball bounces off the block.
By taking e = 1 we get
Substituting the given velocities (and taking right movement as positive):
So the ball bounces back with a horizontal component of velocity of 3v/2.
Alternatively, what happens if e = 0.5, say.
So the ball bounces back with a horizontal component of velocity of 7v/8.
So you can see the value of e is critical.
That's the best I can do with the information given in the problem.
Welcome to the forum.
The x-y system of coordinates was invented by the mathematician Descartes.
But you don't have to use x and y at all. For example to draw a graph of velocity changes with time, you might have 't' across and 'v' up.
You'll find more about coordinates at
Any information about 'coefficient of restitution' ? ie. How 'elastic' are the collisions ?
I could assume it's the same for the block and the floor and hope it will cancel out I suppose.
More thinking: If e = 1 for perfectly elastic collisions, then the ball would come back to its starting position without A moving at all. As A is moving, this suggests that the energy lost during the collisions is exactly the same as the energy imparted to the ball by the block. Hhhmmmm.
Ok. I've thought about it. Let's just think about the vertical motion first.
The initial velocity is zero and the ball accelerates under gravity. When the ball hits the wall the vertical velocity is unaffected because it is parallel to the plain of the block and perpendicular to the motion of the block. So we can just treat the vertical motion without worrying about the block at all.
to get the time to the bottom.
Now the ball will only rise up to the start point if no energy is lost on impact with the ground, so the collision must be perfectly elastic (e=1). This means the time for the ball to rise up is the same as the time to drop, so you can get the total time by doubling.
Now for the horizontal motion. At impact with the block, the velocity is still V and the block is moving at -V/4. We aren't told the mass of the block so no momentum equation is possible. But we can use the elasticity equation. We'll have to assume e=1 again as we aren't told anything else. So
You'll need to be careful with negatives here, but you should be able to get the velocity of the ball after the impact (assume block is still -v/4).
So now you can get the total time for the two horizontal motions and hence get an equation for x in terms of everything else.
It looks like you've got the same woman dancing with two men?
This is what I did (before I researched it).
Get all 30 into the ballroom. Expel everyone else. (except the band of course).
Line the men up against the wall. (So they cannot escape if they don't like their partner )
Man one can have any of 15 partners. Once chosen, man two can choose from 14, man three from 13 etc etc and man fifteen gets the last woman.
Say three men Al, Bob and Colin and three women, Di, Elaine and Fiona.
Al-Di; Bob-Elaine; Colin-Fiona
Al-Di; Bob-Fiona; Colin-Elaine
Al-Elaine; Bob-Di; Colin-Fiona
Al-Elaine; Bob-Fiona; Colin-Di
Al-Fiona; Bob-Di; Colin-Elaine
Al-Fiona; Bob-Elaine; Colin-Di
That's 3 x 2 x 1 = 6 possibilities.
hi jacks and bobbym,
Am I missing something here? Why isn't it 15!
I see you have become a member. Welcome.
I suggest you copy my post and re-post it back with your answers to the ???. Then I can see if you are doing it correctly.
I've not used Management Scientist but I looked it up and it should handle this problem. Have you managed to use it for a simpler linear programming problem?
If yes, then please post that one and your solution. Thanks.
What I am trying to find out is "What can you do already"? So far I am unclear about this.
When there's a lot of text like this, I like to summarise the key information in a table. Here's mine:
Initially I thought there would be three variables, x y and z, for the three grades. But as I created the table, I realised that there are only two choices: how much raw material is given the stage one processing, and then how much of the resulting grade 2 is given the stage two processing. So that fixes just two variables, so I've used x for the first stage processing and y for the second.
So let's say that x is the amount of raw material that is given the first stage process to convert it into grade 1 and grade 2 product. I'll measure x in 100s of litres.
And y is the amount of grade 2 that is given the second stage process to convert it into grade 3 and grade 1 product. I'll measure y in 100s of litres.
So let's now make a model for the stage one process. I'll take x of raw and convert it:
cost = 6x + 3x
time = 3x
grade 1 produced = ????
grade 2 produced = 0.4x
Now let's model the stage two process. I'll take y of the grade 2 and convert it:
cost = y
time = ???
grade 1 produced = 0.4y
grade 3 produced = 0.6y
And the final amounts are now:
grade 1 = ???? + 0.4y
grade 2 = 0.4x - y
grade 3 = ????
total time = 3x + y
total cost = 9x + y
total sale value = ????
I've left a few ??? so you get to put in some of the answers yourself.
One constraint will be that y cannot be bigger than the amount of grade 2 available. ie. y ≤ 0.4x
There will be a time constraint and three constraints due to the maximum amounts that can be made.
The profit will be sales - costs.
Is that enough for you to complete this?
Please post back with your progress.
Welcome to the forum.
Both of these have a bracket that occurs in both terms: (y+1) in the first and (2x-5) in the second question.
So you can draw it out as a common factor:
Note what happens to the term that is squared. To finish this off the square bracket must be simplified.
The second one is similar. Post back your attempt and I'll check it for you.
You are wise to check your answer back in the original question.
-5 is a solution to the quadratic. Does that mean it must be a solution to the log equation ?
First consider this:
A rectangle has a length that is 2 bigger than its width. The area is 24. Find the length and width.
Let the width be x => length = x + 2 so
area = 24 = x(x+2) => x^2 + 2x - 24 = 0 = (x-4)(x+6)
So x = 4 or -6.
Now width = 4 and length = 6, is a reasonable solution. But what about width = -6 and length = -4 ?
As -6 x - 4 = 24, you might accept this uncritically as another answer. But surely we should object to having negative distances.
If you wanted to be bold, you could invent a whole new branch of mathematics where negative distances are allowed. But you'd have trouble drawing this rectangle. It all depends on how the original question was set. Was the questioner expecting only real number solutions ? If so, then reject x = -5. But read on:
The 'laws' for logs of negatives work the same as the laws for logs of positives.
And the definition of logs has been extended to cover this sort of thing:
So, keep up the good work and keep checking those answers.
I found this
just by 'googling' Boolean algebra problems.
If you want to understand calculus, you'll need to know about graphs; straight lines and how to get their gradient**; and then curves and how their gradient varies as x varies. [that's differential calculus ... integral calculus is a summation process that usually follows on from differential]
Some people learn mathematical topics because they need to for a qualification or a job or to help with some other field of study. From what you've said, it sounds like you want to do maths for fun. So the starting point ought to be "What do you enjoy?" Puzzles ? Mathematical Art ? ...? Personally, learning maths is just part of a wider interest that I have, which is to understand how/why things work. eg. Sun dials, why there are exactly 5 Archimedian solids, how you work out bicycle gear ratios, ....
You mention cryptology, economics, game theory, & statistics. Plenty of maths in all of those. Do you want to learn more about those?
**which is why that interactive graph is useful.
Welcome to the forum.
bobbym has already suggested the main maths is fun website, but it is so good it is worth saying it twice. www.mathsisfun.com
Why do I think this is a good place to teach yourself some maths?
(1) It has good indexing and searching, to help you find the right page.
(2) There are cross links on most pages so you can follow up on a point if you want.
(3) The language is clear and, when a new page is made, forum members check it to make sure it has no bugs or confusing words.
(4) The text size and colour make it easier to follow.
(5) There are excellent diagrams.
(6) Many pages have interactive features which greatly help understanding. Here's a few examples:
(7) At the bottom of the page you will find some questions to test your understanding with instant marking.
(8) It's free !
(9) If you cannot find a page on a topic, you can ask for it to be considered as a new page.
(10) It also has a lot of good puzzles.
It's where I go first if I want to look something up.