VCE Methods Question Thread!

[Only Cheating Yourself]

Can someone please solve this for me quickly...thanks

find y

24y-30-4y+2=12y
20y-28=12y
8y=28
y=28/8
y=7/2

Little rusty i think.

[Orb]

24y-30-4y+2=12y
20y-28=12y
8y=28
y=28/8
y=7/2

Little rusty i think.

The dangers of holiday mode

[IndefatigableLover]

24y-30-4y+2=12y
20y-28=12y
8y=28
y=28/8
y=7/2

Little rusty i think.

You got the right answer
Just with your first line, you could have simplified it down with '2' as your common factor (and divide the left side by two to make the right side equal 6y and you pretty much have the same working as the others) just to make things a bit easier

[Only Cheating Yourself]

You got the right answer
Just with your first line, you could have simplified it down with '2' as your common factor (and divide the left side by two to make the right side equal 6y and you pretty much have the same working as the others) just to make things a bit easier

I always forget that… So if the denominators say in this case are 3 and 2, factors of 3,6 and factors 2 2,4,6 so its 6 so multiply by 3 on side and 1 the other side…. Rather than doing 6 and 2 which make 12?

The dangers of holiday mode

Yea umm the thing is i self taught linear solving over the holidays like a month ago…. So its a little worry I've forgotten bits and pieces of it.

Why are you using Null factor Law and CTS and stuff? It's not necessary to the question as the statement does not equal zero....
Just substitute n=4 into the formula.

I don't know…  Its a worded question and its on the null factor chapter, and there were a few questions prior that didn't have a 0 but i added it in and it worked, i suck at worded questions i have no math logic behind it whats so ever.

[Russ]

Combined your posts for you, in future use the edit button pls, rather than posting multiple times

[Only Cheating Yourself]

Combined your posts for you, in future use the edit button pls, rather than posting multiple times

I just messaged another math mod about this…. I didn't know that, so i can't reply to multiple things in a thread?   I really can't afford getting warning like this as I'm on thin ice and that means one more slip up and I'm gone for good which i don't want, also i'm on 40% watch.  Is there  good behaviour thing, where after a period of time being on good behaviour that 40% reduces?  Cause i get nervous every time i get a message thinking I've got another warning etc

Thanks

[Russ]

You can reply to multiple things just not in multiple posts. If you hit the quote button it will quote the post into the quick reply box without deleting what else is there, so you can reply to a half dozen posts that way fairly easily.

You'll be fine with warnings as long as you don't repeat the things you've made mistakes about

[Only Cheating Yourself]

This is frustrating!

The diagram at the right demonstrates the idea of rectangular numbers.
the formula N=w(w+1) give the values of a rectangular based on a shape with a width of w dots.

a.)  which rectangular number has a width of 6?

the diagrams at right are

w=1            w=2                 w=3
N=1             N=6                 N=12

This is my thought process and maybe someone can tell help think a little better?

So i haven't learnt the quadratic formula yet nor has the book covered it yet, so i can only worry about what the book has taught me!  They've given a formula and the question asks which rectangular number has a width of 6.  So its asking for N, theres a formula for that which they give, so don't i plug 6, which is what they're basically saying what is N when the rectangular number has a width of 6, so i plug 6 into the formula.  So the formula is N=w(w+1) so w(w+1)=N which is what we're trying to find right?
n=36w^2+6w, now i look at that and say how can i solve for W, well i need a constant don't i?  So i can use the null factor law, which basically solves for x… in this case w.  So i need a constant how can i get that constant?

Thats my thought process, am i on the right track or? 

[M_BONG]

Can anyone help me double check the answers i got in this question please?

Find the area bounded by the curve y= sin(x) - cos(x)  x € (0, pi) between the x axis, 0 and pi.


Answer I got is root(2) square units.

Maths Quest says it's 2 root (2) square units.

[brightsky]

area = - int^(pi/4)_(0) (sinx - cosx) dx + int^(pi)_(pi/4) (sinx - cosx) dx
= - [-cosx - sinx]^(pi/4)_(0) + [-cosx - sinx]^(pi)_(pi/4)
= - (-sqrt(2) + 1) + (1 + sqrt(2))
= sqrt(2) - 1 + 1 + sqrt(2)
= 2sqrt(2) units^2

[Only Cheating Yourself]

Can anyone help me on the question i posted above?

Thanks

[hobbitle]

But I'm pretty sure you just substitute in '6' for 'w' and solve for N.
ie.
N = 6(6 + 1)
N = 42.

[hobbitle]

n=36w^2+6w

You were right in until this point. I have no idea where you got this formula from... But yeah you just plug in '6' for 'w' and solve for N. Sorry mods for repeated posts, I'm on my phone and it's not possible to multiquote.

[Only Cheating Yourself]

You were right in until this point. I have no idea where you got this formula from... But yeah you just plug in '6' for 'w' and solve for N. Sorry mods for repeated posts, I'm on my phone and it's not possible to multiquote.

w(w+1) plug in 6 isn't 6w time 6 w which makes 6w^2 or do you totally replace w with 6?

[hobbitle]

You totally replace w with 6.