I'm not saying that is what happened here but sometimes new posters put up their homework and then panic because they worry their teacher will see they haven't done the question themselves. A few years ago someone posted a question that was part of a competition. The rules stated that competitors should not seek help on the internet. The question master did a search and uncovered the post. Hhhm.
When someone asks for help with a problem I try to give the minimum that will enable the poster to fill in the gaps themselves. That way they get maximum benefit from the 'help'. Of course, it is tricky knowing what 'minimum' means for any particular poster which is why the 'sticky' at the head of the help me section says (amongst other things) "Tell us what level you are at. It will help avoid replies that are too easy or too hard for you to understand." and "make it clear what kind of help you want."
Welcome to the forum.
The swimming pool question is impossible without a diagram. We need to know what shapes it is made from. If you have a diagram please post it. eg. Upload it to imgur.com and then insert the BBCode into a post here. If you cannot do that then how about just describing it.
eg. The ends are semicircles when seen from above and these are joined by a middle section that is a rectangle. The measurements are ............
Below the water line the depth is not constant. At the shallow end it is ....... and then it gets steadily deeper towards the other end where the depth is .......
If you can do something like that I'll make you a picture and say how I would do the question.
Thanks for the explanation. I have tried chrome and it works as expected. Sadly my ears don't . My wife has been telling me to go and get a hearing check. I think you've just prompted me into doing exactly that. An additional use for the page that you may not have planned . Thanks.
Limit on fuel means you must wait for the ship to get close enough and then 2 hours flight time, hospital to ship.
So if you wait for the ship to do 80 miles (8 hours) and then the helicopter sets off, 2 hours later they meet at a distance of 300 miles from the hospital and 100 miles from the ship's original position.
If the 'mayday' is a 5.30am then the helicopter will arrive at the ship at 3.30pm on that day.
Welcome to the forum.
Yes, you can tell.
If a quadratic looks like this:
then you can divide by a to get
If this factorises as a perfect (complete) square then it will look like this:
so by comparing coefficients
You may wish to simplify this by cancelling an 'a'.
So mathematicians invented index notation to apply to positive whole numbers like this
When y is multiplied by itself 'n' times:
This definition results in three laws of indices:
n and m are positive whole numbers (with n greater than m):
But mathematicians then wondered if there is a sensible way to extend the definition so that n and m can be any numbers.
If you specify that the three laws must still be obeyed then you can work out sensible definitions for y^n when n is not a positive whole number. For example, what about if n is zero?
using the first law.
So it is sensible to define y^0 to be 1 for all y.
Then if n is a negative whole number:
using the first law and what we have just learnt about a power of zero.
So define as follows:
This is now consistent with the first law as we have
You can continue like this is find sensible definitions for fractional powers and so on.
hi Zeeshan 01
Limits are used to show what a sequence of function calculations comes to as 'x' approaches a particular value. So for f(x) = 3 + x and x approaching 2 we might try:
f(1.9) = 3 + 1.9 = 4.9
f(1.99) = 3 + 1.99 = 4.99
f(1.999) = 3 + 1.999 = 4.999
and so on.
You can see that as x gets closer to 2, f gets closer to 5.
The same limits applies if x approaches 2 from above:
f(2.1) = 3 + 2.1 = 5.1
f(2.01) = 3 + 2.01 = 5.01
f(2.001) = 3 + 2.001 = 5.001
With your example anyone can calculate f when x = 2 anyway, so there is nothing remarkable about the limit.
But sometimes it is not possible to calculate a function at a certain value eg. f(x) = tan(x) where x is measured in degrees cannot be evaluated at x = 90.
By trying 89.9, 89.99, 89.999 and so on you will discover that the limit is infinity.
But try 90.1, 90.01, 90.001 etc and you will find the limit is minus infinity. So left and right limits are not always the same.
There's a starter page on limits here: http://www.mathsisfun.com/calculus/limits.html
Post #5. Thanks for finishing that off. I was away from my laptop and trying to do this on a kindle without any pencil and paper. So I just posted what I could visualise. I knew a variable could be eliminated but hoped the poster would be able to do that step alone.
Turns out to be easier than I thought.
Work back from the available fuel. 6600 to start but 1200 must be kept as reserve and half hour = 600 for the transfer so the helicopter has 6600 - 1200 - 600 = 4800 for its journey. It must go out and back so it can travel to a point at sea 2400 pounds of fuel away from the hospital = 2 hours of flight time.
In 2 hours the ship will cover 20 miles and the helicopter 300 miles so they won't meet. That's why the helicopter must wait while the ship gets nearer before taking off.
So wait while the ship travels 80 miles, then set off and meet at a point 300 miles from the hospital.
I'm assuming the two must be stationary during the transfer. You could shave a bit more off the time if the helicopter can meet the ship, turn around and match its speed while both travel towards the land. Then once the passenger is aboard the helicopter, it is nearer the hospital so won't take 2 hours to get there. Can you re-calculate using this interpretation? post back if you want me to try it.
For Q2, you can split the triangle down the middle to create a right angle.
For Q3, there is a right angled triangle, so call the opposite y, and the adjacent x. Form an equation from the given trig. equation and another from Pythagoras. and then solve for x and y.
Hope that works ... I am trying to do this on a kindle.
I'll have a go at trying to explain. Let's say we have a 3 dimensional space of all possible vectors. I'll write my vectors across the line to make typing easier. The basis (1,0,0); (0,1,0);(0,0,1) is said to span the space because you can construct a linear combination from the basis to make any possible vector in the space.
eg. (2,3,4) = 2(1,0,0) + 3(0,1,0) + 4(0,0,1)
The basis doesn't have to be that simple. It could be (1,1,0); (0,0,1); (1,-1,0)
In that case (2,3,4) = 2.5(1,1,0) + 4(0,0,1) -0.5(1,-1,0)
check 2.5(1,1,0) + 4(0,0,1) -0.5(1,-1,0) = (2.5,2.5,0) + (0,0,4) - (0.5,-0.5,0) = (2.5 + 0 -0.5, 2.5 + 0 +0.5, 0 + 4 + 0) = (2,3,4) as required.
To make a basis there must be exactly as many vectors as the size of the space ... so 3D needs 3 vectors and 2D needs 2 vectors. But it is not enough just to have sufficient vectors. None must be dependant on the others ... or to put it the other way round, the basis vectors must be independent of each other.
Let's see what happens if they are not independent.
Say I choose (1,0,0); (0,1,1); and (1,1,1) Note (1,0,0) + (0,1,1) = (1,1,1) so the third is a linear combination of the first two.
Now let's try to make a linear combination to make (2,3,4)
a(1,0,0) + b(0,1,1) + c(1,1,1) = (2,3,4) Can I find a, b and c ?
a+c = 2
b+c = 3
b+c = 4
Clearly the second and third are inconsistent with each other. b+c cannot be 3 and also 4. So no solution exists. If you tried plotting lines you would find these two equations lead to parallel lines and so would not cross.
I'll change the third vector to one that is independent of the other two. The first two combined can make any vectors of the form (a, b, b) so I only need to choose my third vector so that its 'y' and 'z' components are different. eg. (1,2,3)
So I should be able to make a linear combination out of these.
a(1,0,0) + b(0,1,1) + c(1,2,3) = (2,3,4) Can I find a, b and c ?
a+c = 2
b+2c = 3
b+ 3c = 4
Subtracting the second from the third we get c = 1, which means b = 1, and finally a = 1 from the first equation.
check 1(1,0,0) + 1(0,1,1) + 1(1,2,3) = (1+1, 1+2, 1 +3) = (2,3,4)
Hope this helps. Post again if you need more help.
Which is the tens digit of the sum of the first 30 digits in the sequence?
eg. In this number, 2468, I would say it has four digits: 2 then 4 then 6 then 8. So I'm guessing you mean the first 30 numbers in the sequence.
To get an answer: All 30 have a units digit of 1. 29 of them also have a tens digit of 1. What happens for the hundreds, thousands etc. doesn't affect the answer. So compute 30 x 1 + 29 x 10 and look at the tens digit of your answer.
Q1 ? The key words here are 'reaches' and then 'clears'. The front end reaches but the rear end clears. If the train is 'd' long and the tunnel is 9d long then the rear end of the train has travelled d + 9d when it clears the tunnel. So divide by 10 not 9.
If x - y + 2z, and x+z = y + t, and y + t+z, how many z's = x?
Do you mean
x-y+2z = x + z = y + t = y + t + z
No. That's silly. I think the EQUALS and AND signs are misplaced here.
a athelete running 100 meters or 200 meters does not reach top speed until 40 miles
Do you mean "An athlete running either 100 metres or 200 metres reaches a top speed after he has run 40 metres." In which case are we to assume that he accelerates at a constant rate until this moment? Alternatively, it would be possible (but it's a tougher calculation) to have a steadily dropping rate of acceleration until it is zero after 40 metres.
In a 200 meter race a certain runner's time is 5.4 secs and his time for 100 meter s is 10 secs.
That's clever I'm hoping this has a typo because it's a strange running model otherwise, and a new Olympic record!
what would be his time for 200 meters?
Eh! Isn't it 5.4 secs ? You seem to have a serious gremlin infestation in this question.
Welcome to the forum.
Do you mean help from the forum or from the main site. Some members discover the former long before the latter. In case that applies to you go here http://www.mathsisfun.com/
You'll find lots to help you there!!!!! eg.http://www.mathsisfun.com/numbers/number-line-zoom.html
Thanks. I stand corrected. I'll re-try my simplification. Let's hope it comes out nicely with this correction
Note: This change will affect alpha squared + beta squared as well.
I should have said:
Working as before with these equations I got this:
There's a page of algebraic simplification involved and I often slip up with things like that so this may not be right. I'll have a break and then try to check it.
Let's say they meet after T mins at a point x around the track. Also A has completed n complete laps (+ x); B m laps; and C p laps. Then
Eliminate x by subtractions and we get
Clearly p > m > n (and all are whole numbers) so inspection shows p = 3; m = 2; n = 1 will work.