VCE Methods Question Thread!

[kinslayer]

0.06/12 = 0.06/(6*2) = 0.01/2 = 0.005

[IndefatigableLover]

How would you do 0.06 divided 12 by hand?

Adding onto the other suggestions (more so if they want your answer in exact form rather than in decimal form):








(through simplification)

[cosine]

Can someone give me a quick run down on solving linear equations for unique (infinite?) and no solutions? I have no idea how to do it and what it means.

thank you!

[Apink!]

Hi!
I need help with this question. I have a SAC tomorrow, and I am freaking out

The matrix that describes the composition of mappings
- dilation of factor 5 from the x-axis
- reflection in the line y=x
-reflection in the y-axis

In general, I am confused about questions like this (using matrices!)
Please help

[knightrider]

Consider the following transformations:


It's 0.06/12.
Done. Exact form. (of course, normally we take out the decimals, so I'd multiply top and bottom by 100)

0.06/12 = 0.06/(6*2) = 0.01/2 = 0.005

Adding onto the other suggestions (more so if they want your answer in exact form rather than in decimal form):








(through simplification)

Thanks so much guys 

[Eiffel]

need a bit of help here:

the line with equation y = 2x - k where k is a real number, intersects the parabola y = x^2+2x-3 at two distinct points.

I've bee through countless questions like this so why i did was

y = y
x^2 + 2x - 3 = 2x - k
x^2 - 3+k = 0

b^2-4ac > 0
-4(1)(-3+k) > 0
-4(-3+k)>0
12-4k>0
k< 3, answer has k > 3?

[Eiffel]

y = p/x
xy = p
xy/p = 1

is this correct

[kinslayer]

need a bit of help here:

the line with equation y = 2x - k where k is a real number, intersects the parabola y = x^2+2x-3 at two distinct points.

I've bee through countless questions like this so why i did was

y = y
x^2 + 2x - 3 = 2x - k
x^2 - 3+k = 0

b^2-4ac > 0
-4(1)(-3+k) > 0
-4(-3+k)>0
12-4k>0
k< 3, answer has k > 3?

k > 3 is right, is there a typo in the question?

y = p/x
xy = p
xy/p = 1

is this correct

You divided by x in the first line and p in the third line. So it is correct but only if neither one of x and p is equal to zero.

[Eiffel]

k > 3 is right, is there a typo in the question?

You divided by x in the first line and p in the third line. So it is correct but only if neither one of x and p is equal to zero.

-4(-3+k)>0
12-4k>0

then -4k > -12
k< 3, how is this wrong?


....

the question was
loge(y) = loge(p) - loge(x)
loge(y) = loge(p/x)
y = p/x
xy = p
xy/p - 1 = 0???

cheers

[Stevensmay]

-4(-3+k)>0
12-4k>0

then -4k > -12
k< 3, how is this wrong?

Whenever you multiply/divide by a negative number you need to change the direction of the sign. So from -4k > -12 you divide by negative 4 so you must change the sign. This gives k > 3.

Consider -2 < 2. If we multiply both sides (without changing the sign) by -2 we get 4 < -4 which is quite clearly not true. So we flip the sign to get 4 > -4.

[Eiffel]

yeah i have havent i haha....?

-4k > -12
k< 3 .... i'm saying how is this wrong? if i wasnt to change the sign then k > 3 which is wrong.

answer has gone from
-4k > -12
to
k > 3 and not k < 3

 

Whenever you multiply/divide by a negative number you need to change the direction of the sign. So from -4k > -12 you divide by negative 4 so you must change the sign. This gives k > 3.

how have you changed the sign?

[StupidProdigy]

How do you solve |2x+1|=x+1 graphically?? Thanks

[knightrider]

why is ?

[psyxwar]

why is ?

Using our exponent laws:

[cosine]

Is a 50 still plausible if I lose about 5 marks on a sac out of 60? Because according to some resources, you can only lose 'a few' marks throughout the year. Worried :3