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#1 Re: Help Me ! » Sequences and Series » 2007-01-31 00:47:37

I think I should have made a change in this step:
I've equated

as

Which is just

#2 Re: Help Me ! » Sequences and Series » 2007-01-30 11:28:18

So they'd definately accept my answer?  What I don't understand is - why would they teach me this method that I use on all of their questions, then put the answer in a different format?! It's a crazy world we live in :S

#3 Re: Help Me ! » Sequences and Series » 2007-01-30 08:38:27

I see this, is it then that the answer the book suggests is a simplified version of my answer?  If so, what is the common method to simplify?

#4 Help Me ! » Sequences and Series » 2007-01-30 08:30:44

RickyOswaldIOW
Replies: 7

Write the following series in ∑ notation:

This is a very simple progression but the technique I am using does not seem to yield the right answer, which is:

Here is my method;

Work out the common difference, I subtract the first in the series from the second:

So the difference is 1 -> 1r.  Next I go back one term before the start of the series:

Now I add these two terms together for the general term:

As you can see, I've already gone wrong! But I will continue...  Find the values of r for the first and last terms:

and:

Now I write down the ∑ notation:

This method works fine for the other questions, where am I going wrong?

#5 Re: Help Me ! » Constant / Fraction » 2007-01-28 23:27:15

Okay, that's brilliant and I understand how to do it! Thanks.

Could you have a look at my working out here? I'm not sure where I have gone wrong:

Find the points where the given line meets the given curve:

2x - 5y + 17 = 0   meets     xy = 6

Firstly I get y on its own:

5y = 2x + 17                       y = 6/x

y = 2/5x + 17/5

Now I can place the two side by side:

2/5x + 17/5 = 6/x

Now I balance and collect all the terms:

2/5x^2 + 17/5x = 6

2x^2 + 17x = 30

2x^2 + 17x - 30 = 0

Now I find two numbers whose sum is 17 and product is 2*-30 (or -60).  These two numbers are +20 and -3, I re-write the quadratic:

2x^2 - 20x + 3x - 30 = 0

Now I take out the common factor of the first two numbers, then the second two numbers making sure that both sets of brackets are the same:

2x^2 - 20x         + 3x - 30        = 0
2x(x-10)             + 3(x-10)       = 0

This leaves me with:
(2x+3)(x-10)

So x=10 or x=-3/2

I substitute these values into y=6/x to give me

y=6/10   and     y=6/(-3/2)=-4

So the answer should be that the line intersects the curve at (10, 6/10) and (-3/2, -4).  My book gives the answers with the signs reversed!:

(3/2, 4) and (-10, -3/5)

Whose signs are backwards?

Edit:

These two numbers are +20 and -3, I re-write the quadratic:

2x^2 - 20x + 3x - 30 = 0

My signs are wrong Thanks again!

#6 Help Me ! » Constant / Fraction » 2007-01-28 22:53:26

RickyOswaldIOW
Replies: 3

What is the common method to work out a fraction with three stages?

I have 6/x and I've worked out x=-3/2 :
6/(-3/2)

I may have the wrong value for x, but I'd still like to know a good method for doing this

#7 Re: Help Me ! » Intersection of a line and curve. » 2007-01-25 08:44:43

Aha! So I just substitute the value of y into the first equation?  Thankyou luca

#8 Help Me ! » Intersection of a line and curve. » 2007-01-25 08:40:02

RickyOswaldIOW
Replies: 4

Hello all!  I've got a small problem here, I have to find the intersection of a line and curve, I can do this just fine on most of the questions but I'm having difficulty on this one:

2x + 3y = 14 meets xy = 4

Obviously I have to re-arrange the first to get y = 14/3 - 2/3x,  it is the second that I am having difficulty with.

How do I re-arrange xy = 4 to be a quadratic equation?  Since I didn't take my GCSE maths and went straight on to Advanced level, there are some gaps in my knowledge so if you could explain how I can re-arrange this and other similar equations it'd be much appreciated!

So far I've taken it to the step: y = 4/x

Am I on the right track?

#9 Re: Help Me ! » Intersection of two lines » 2007-01-20 07:00:35

3x - 10 = 4 - 1/2x
7/2x = 14
7x = 28
x = 4!

Thanks anyway

#10 Help Me ! » Intersection of two lines » 2007-01-20 06:57:28

RickyOswaldIOW
Replies: 1

Find the points of intersection of the following pair of lines:
2x + 4y = 16
y = 3x - 10

Firstly I get the value of y from the first line

2x + 4y = 16
4y = 16 - 2x
y = 4 - 1/2x

Now I equate the two expressions for y

3x - 10 = 4 - 1/2x
5/2x = 14
5x = 28
x = 28/5

I think this is where I have made the mistake, but I will go on

Now we can use the value of y to find the value of x

y = 3(28/5) - 10
y = 84/5 - 10 = 134/5

Thus the point of intersection is (not) (134/5, 28/5)...

#11 Re: Jokes » its a bit funny.. hehe » 2007-01-07 12:13:10

Q: Why are pirates pirates?

A: Because they ARRRR!

#12 Re: Jokes » joke about pi » 2007-01-07 12:10:15

not unless you pwned him

#13 Re: Help Me ! » More Curves » 2006-09-13 03:01:17

I still don't understand this.  So far the book has covered how to sketch the graphs of y=x^2, y=x^3 and y=1/x and how to translate and scale them.  It also taught us how to find the vertex point and the points where it intersects the axis.  If the question is asking me for the y value at x=-1 then, following what the book has taught so far, I don't know how to do it.
I can work the answer out because I am past this point in the book, I don't know how to work it out using any of the methods the book has taught so far:
Factorising and
Completing the Square

Any ideas?

#14 Re: Jokes » Not Suitable For Kids!! » 2006-09-11 13:26:47

Devanté's joke is better.

#15 Re: Help Me ! » More Curves » 2006-09-11 13:15:23

Am I understanding the equation correctly? 1(1-x) is just 1-x (Plot of 1(1-x))

It's a curve, not a straight line.  I don't have the book to hand but it must be x(1-x), typo on my part .  The curve is a curve of x².
x(1-x)
-x² + x

If you plot this curve on an axis you can read out the minimum and maximum co-ords of the points between x = -1 and x = 1 i.e. (-1, y), (1, y) and the vertex (1/2, 1/4).

#16 Re: Help Me ! » More Curves » 2006-09-11 13:10:43

EdExcel, NEC - I study from home.

#17 Re: Help Me ! » More Curves » 2006-09-10 11:10:52

Is it asking for the maximum and minimum points of the curve between x = -1 and x = 1 on the axis?  i.e. the maximum point of the curve is (1/2, 1/4) and then the two corresponding points at (-1, y) and (1, y)?

#18 Re: Help Me ! » More Curves » 2006-09-10 11:01:06

Unfortunately, A-level maths C1 doesn't do "interesting"

#19 Help Me ! » More Curves » 2006-09-10 08:31:30

RickyOswaldIOW
Replies: 10

Sketch the graph of the curve with the equation y = 1(1 - x). Determine the greatest and the least values of y when -1 ≤ x ≤ 1.

I can sketch this curve just fine, but what is meant by the last sentance?  Is it asking for a set of co-ords on the curve where x = -1 and x = 1?

#20 Re: Help Me ! » Where does the curve meet the axis? » 2006-09-10 07:38:20

Brilliant, thank you

#21 Help Me ! » Where does the curve meet the axis? » 2006-09-10 06:30:05

RickyOswaldIOW
Replies: 3

sketch the curve y= 9 - (x - 2)², showing the coordinates of the points at which the graph meets to coordinate axes.

I have produced this sketch, the maximum point (2,9) and it crosses the y axis at (0,5).  I know to find the intersection of the x axis I must put y=0 i.e.  y = 9 - (x - 2)² = 0.  I can't remember how to solve this and it's been puzzling me all last night and today
Firstly I multiply out the brackets to give
-(x - 2)² - 9
-x² + 4x + 4 - 9
-x² + 4x - 5
So,
-x² + 4x - 5 = 0
-x² + 4x = 5
Where do I go from here?

#22 Re: Help Me ! » Sector areas. » 2006-07-17 08:08:45

Welcome to the forum DASET

#23 Re: Help Me ! » Sector areas. » 2006-07-17 04:02:25

I have not been taught this method yet but it may appear soon.  Thanks a lot for the help though

#24 Re: Help Me ! » Sector areas. » 2006-07-16 22:12:41

One statement you said has 3 variables and another only 2 variables

a² = b² + c² -2abCosθ is the genral formula but since my triangle was already labled I just had to rename the variables.  it could have been x, y, z or anything really.

ou choose a round about way of computing the angle and lost accuracy due to the lack of precision in your conversion to radians

Do you mean in the step where I write "θ = 0.4"?  I probably should not have rounded this figure up to 0.4, I'll retry the sum without rounding the value of theta.

#25 Help Me ! » Sector areas. » 2006-07-14 23:51:36

RickyOswaldIOW
Replies: 6

I have the following triangle from which I have to work out the area of the shaded part.

I am firstly working out the area of the triangle:
triangle area = (c*b)/2 = (5*12)/2 = 30

I know the formula to work out the area of the sector is:
(r²θ)/2

To work out θ I use:
a² = b² + c² -2bcCosθ

I can work out the length of a using:
a² = b² + c²  ->  a² = 12² + 5²  ->  a² = 169  ->
a = √169 = 13

Thus:
c² = b² + a² -2baCosθ
5² = 12² + 13² - 2(12)(13)Cosθ
25 = 313 - 312Cosθ
312Cosθ = 288
Cosθ = 288/312 = 0.9230...
θ = 0.4

And so the area of the segment is:
(r²θ)/2 = (12² * 0.4)/2 = 28.8

If I subtract this value from the area of the triangle I should be left with the area of the shaded part:
30 - 28.8 = 1.2

My book states the answer as 1.58?!