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Welcome!

For part (a), what happens when you multiply that matrix by each of those vectors?

For part (b), you'll have been able to find some number (called a 'scalar multiple') such that , and likewise for the others. This then tells you that . Use this to write down an expression for and therefore .For part (c), trial and error is easiest here: try to find a way of making 10 from a linear combination of and . Then see if that same linear combination works for the other two vector components.For part (d), you can use your linear combination found in part (c) to replace that vector with something that looks like . You can then split the matrix multiplication into three parts: can you see how?Welcome, Xavier!

Hi Lasson,

Can you please check your other percentages thread?

Hi,

You might find this link useful:

Hi Alexandr2,

I responded to your other post -- please have a look when you have time.

One of them secured 9 marks more than the other

Let me know if you get stuck.

Hi Russia,

Have you considered registering an account with us? You can do this by .

The old price of the coat is $20, and the new price is $15. That means the price has reduced by $5 (since 20 - 15 = 5). So now you just need to find out how much percent $5 represents as a fraction of the old price. This is:

Please let us know if you need any further help!

you have this instead:

Another way of writing this is:

Now, let's say you wanted to solve this system for and . You'd need to find the inverse of , right? But for that inverse to exist, the determinant can't be equal to 0. In other words, for that thing to have a solution, you must have:i.e.

which tells you that isn't a multiple of . Remember, saying that and are scalar multiples of each other just means that you can find some real number such that , or in other words:i.e.

Now, for the forwards direction (any vector can be expressed as => and aren't multiples of each other), try proving the contrapositive, i.e. show that if for some real number , then not every vector can be expressed in the form . Let me know if you have any more questions!No worries -- let's look at part (a) first.

!nval!d_us3rnam3 wrote:

Okay, let's take any two-dimensional vector, say, . We want to know: can we find real numbers and so that:(a) Show that any two-dimensional vector can be expressed in the form

where and are real numbers.

If we 'multiply out' the left-hand side, we get:

Now, we can add two vectors just by adding the matching components, so that:

In other words, we want to find real numbers and so that:which is exactly the same as solving the pair of simultaneous equations:

Remember, we're solving for and here. (Just pretend that and are any old real numbers.)Let me know if this makes sense -- happy to explain anything further if you need more help.

Hi !nval!d_us3rnam3,

Thanks for your post -- I fixed your LaTeX.

For part (a), suppose you've got some vector in . Then, you've just got to solve this pair of simultaneous equations for and :Does that make sense? (Let me know if anything sounds confusing -- happy to help.)

For part (b), suppose that instead of and you have and . This gives you the system:What sorts of conditions do you need here for that to have a solution?

Thanks phrontister, that work pretty well. Hard to believe it has been two whole years.

Anthony Lahmann wrote:

I will integrate 1/sqrt(1-x^2) by u-substitution. Here's how I did it:

After we did the u-substitution, we end up with the exact same integral, but with a negative in the front. What happened?

Your issue is here:

The correct implication is:

How many breakdowns occur in total?

Fingers crossed!

Monox D. I-Fly wrote:

And here I thought that A' in the context of sets meant the complement of A.

It can indeed -- the notation can be quite varied!

A' is read as 'A dash' in British English and 'A prime' in American English, although at my university 'A prime' was far more common, probably due to American influence.

In *Littlewood's Miscellany*, Littlewood jokes about this notation used in the context of sets (in point-set topology, A' is the set of all limit points of A, so that A' is called the *derived set* of A).

John E. Littlewood wrote:

I have had occasion to read aloud the phrase "where E' is any dashed (i.e. derived) set". It is necessary to place the stress with care.

Hi Amartyanil,

Yes: try looking at that equation modulo 2 and modulo 3. From there you can deduce that *x* is a multiple of 2 and that *y* is a multiple of 3, which allows you to reduce that equation into something much simpler!

Welcome to the forum, Kamov!

Hi Math 1122,

Welcome! Why not register an account with us?

You can start by using the identity

You'll also need (some, if not all of) these facts:

Hi Bob,

I made a video explanation here, if Zeeshan 01 would like to have a look.

Set

then use the trig identity

Hi segfault,

Welcome to the forum!

Suppose that there are some values x and y for which ∛(7 + 5√2) = x + y√2. What happens if you cube both sides of that equation?