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Hmmm...The famed Wolfram Alpha, which uses mathematica, returns no solutions to both questions. There is a definite error.
These questions are extremely odd, maybe we are both missing something, but I simply cannot see it, writing the equations in terms of x necessarily yields 0 = an integer.
Is it possible that the question is, in fact:
?
Well if we just look at the problem quickly we can see that there is -12x on both sides which must cancel. There is definitely an error here, are you sure that you have copied the question out correctly?
Oh dear, well good luck! I'm not sure that I think that much of this instructor. I get -251 = 0 as well and I can't see anything wrong with your algebra - although, if you don't mind me saying, you could probably have made your life a little easier in places. Perhaps your instructor's question was incorrect. Unless anybody can see our mistake?
Of course I'll be gentle. Now why has the
in the original post become
?Hi Charlie, what have you tried so far?
Be careful, remember that there is a minus outside your brackets so you need to subtract everything within the brackets from that which is outside thereof.
Hi there,
No I don't think that that is possible in this case. It is, however, perfectly acceptable to have As on both sides, simply rearrange it such that you have it in that form.
Certainly it's possible, however, you will have As on both sides.
Can I offer a suggestion regarding the typesetting of matrices. The array environment is very useful, however, there is a dedicated environment which some may find useful, simply use:
\begin{φmatrix}
\end{φmatrix}
Where φ is a variable. pmatrix will produce a matrix with parentheses, or standard brackets, bmatrix one with square brackets, Bmatrix one with braces, or curly brackets, vmatrix one with straight lines and Vmatrix one with double straight lines. Simply using matrix will create a simple array, with no brackets of any kind. As before entries are separated with & and \\ will drop a line. Note adding \\[φpt] will leave a gap of φ point. Seeing as LaTeX is, by default, usually 12pt, I suggest \\[12pt].
To exemplify
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
Produces
\begin{vmatrix}
a & b \\
c & d
\end{vmatrix}
Produces
etc.
Hmmm, I haven't quite multiplied by A, I didn't find that helpful, so I divided by
, which is of course equivalentSince
is defined as
Thanks a lot, once again, Bob, your help has been invaluable, I was completely stumped as to what to do and probably would never have spotted that link .
I was wondering what you think of this solution:
Hmmm...that's a really good way of thinking about it.
Since
Therefore, by multiplying both sides by
I'm really not sure what to do here, I can't multiply both sides by
However, can I use the reflexivity of addition to say that
Since
Therefore
Hi guys,
Here's an interesting little question.
Given that
prove that
Since
But I don't really know how to divide or factorise matrices, since multiplication is not commutative, so I don't really understand why its
And not
Hi,
Thanks, I think I've got it, this is the figure as given in the question.
Thanks .
Hi Bob, thanks again. Yes the side BC is completely flat against the plane, I have drawn the image in a simple picture editor and have it as a .png file on my computer, but I'm not sure how I would go about uploading it. Part (b) is a style of question which is asking about what happens as you increase the angle of plane. Obviously there will come a point, assuming that the lamina doesn't slide down the plane under gravity, at which the mass acts outside of the lamina, which, obviously, will cause the lamina to roll down the surface. This question is asking for the greatest value of theta for which the lamina remains in equilibrium - i.e. the point at which the mass is acting through the right-angle at the point B. As I'm sure will now be clear to you, all that we need to do is construct a right-angled triangle of base 3 and height 4 (i.e. a right angled triangle from the centre of mass to the plane) and angle theta at the centre of mass. Then, of course, it is a simple tan calculation.
Thanks Bob!
Yes it does have a part (b):
Assuming that the plane is rough enough to prevent the lamina from slipping,
(b) find in degrees, correct to 1 decimal place, the maximum value of θ for which the lamina remains in equilibrium.
But even this is only worth 4 marks.
Hi again,
I'm very confused about one question, specifically:
Figure 1 shows a uniform triangular lamina ABC placed with edge BC along the line of greatest slope of a plane inclined at an angle θ to the horizontal. The lengths AC and BC are 15cm and 9cm respectively and
is a right angle.(a) Find the distance of the centre of mass of the lamina from
(i) AB,
(ii) BC.
Now, the centre of mass of a right-angled triangle, unless I am very much mistaken is the point which is one third of the way along each of the sides which form the right angle. Therefore, I would have thought that the answer to (i) is:
cmand to (ii) is:
cmHowever, the whole of part (a) is worth 6 marks. Am I missing something?
Ahhh, I see, thanks very much Soroban, I had made a mistake in the differentiation and hence could only get one positive value of t. Thanks again!:)
Hi,
I was wondering if anybody could help me out by reminding me how to find the times at which a particle comes instantaneously to rest - I can't seem to remember how to do it. I think one has to let the velocity = 0, or perhaps the acceleration, but neither of these seem to get me the right answer, as I can only get one value of t and the question implies that there should be more.
In case it is helpful, and so as to make it clear what I am talking about, here is the question:
A particle moves along a straight horizontal track such that its displacement s metres, from a fixed point O on the line after t seconds is given by
(a) Find the values of t for which the particle is at O.
Unless I'm very much mistaken the solutions to part a are
(b) Find the values of t at which the particle comes instantaneously to rest.
Thanks!
Ooooh thank you Bob Bundy, I believe that is the method which I'm using, but it's good to see it stated so elegantly and concisely, I feel that I have a much better understanding now - thanks a lot
Hi Au101;
Yes, Bob gave the answer away. Since you were translating by B that is why it delivers the clockwise rotational matrix. Seem to make sense. As an exercise translate some points using A. They should go counterclockwise. That is what you want.
Note to self: always read the textbook thoroughly before embarking upon the exercise
That is where I disagree. Play as much as you like with math. We are trying to bring it back to being an experimental science. The way it used to be!
Hehe:) I admire your outlook on maths very much and whilst I quite agree it appears that the question before consisted of 7 invertible transformations meaning that I had done plenty of practise of the wrong thing! As much as I hate textbooks, if I have to read them, I think I should probably make sure that I don't misread them and learn the wrong thing!
Thanks again to both of you!
Thanks for your help Bobbym, I have had a look and that matrix is definitely correct.
It seems, however, that Bob Bundy is quite right, thanks very much, I hadn't realised that I'd been doing it wrong all the time!
Thanks very much to both of you, that should have it fixed.
Note to self: always read the textbook thoroughly before embarking upon the exercise
Hi guys, I'm having trouble with a particular calculation. I know that a reflection in the y-axis followed by a reflection in the line y = -x is equivalent to a rotation of 90 degrees anticlockwise about (0,0). A friend asked me to prove it and even though I can quite plainly see that it is by drawing a graph I simply cannot get the matrix algebra to add up.
I know that a reflection in the y-axis can be denoted by the matrix
And I'm almost certain that a reflection in the line y = -x can be denoted by the matrix
However the product of these two matrices comes out at, unless I'm very much mistaken
When it should be
Can anybody see my mistake? Thanks:)