VCE Methods Question Thread!

[Phy124]

Solve , correct to three decimal places.

I would have thought that equation would have an asymptote at y=0 and therefore solving for x=0 (assuming that's what you meant) would give no solution.

Considering it says "correct the three decimal places" you should be able to use your calculator, but I still can't see you getting a solution.

[jasoN-]

I'm feeling really ill, so my latex might be off.
* = multiplied

yeah i know, but what does that equation equal? there's nothing to solve
e.g. y = 2*2^(-2x)??

[Kanon]

2*2^(-2x) = what?

also for the logs thing, doesn't matter which way, but i prefer keeping everything positive if possible:
ln(x) - 2ln(5) = 2
then ln(x/25) = 2
x = 25e^2

I'm feeling really ill, so my latex might be off.
* = multiplied

EDIT:
I'm not sure if i'm right or not, but I did:

[Kanon]

I'm feeling really ill, so my latex might be off.
* = multiplied

yeah i know, but what does that equation equal? there's nothing to solve
e.g. y = 2*2^(-2x)??

Oh my god, i am a freaking idiot haha, I didn't notice this.
It must be the colddd but 2 * 2^(-2x) = 2002

I am so sorry haha

[jasoN-]

yup you are right

[ICECOLD]

hi. new to this, is this where I ask a question?

Given X~N(32,16). Fish which are less than 27cm are considered to be undersized.

Find a) Pr fish is undersized?

I got this as 0.1057

b) the expected number of fish that a fisherman could take home if he catches 20fish, ruling out the undersized fish.

I did 20x0.1057 to get 2.2197 fish undersized.

Do I do 20-2 or 20-3. Don't know how to round.

thanks.

can anyone pls help?

[Kaille]

help!

transform the graph y = 1/x^2

1. dilation of factor 2 from the x axis
2. dilation of factor 3 from the y axis
3. reflection in the x-axis
4. translation of 3 units in the positive direction of the y axis

i get y = 18/x^2 +6, but the answer says y= 18/x^2 +3....

[brightsky]

y=1/x^2
y=2/x^2
y=2/(x/3)^2 = 18/x^2
y = -18/x^2
y = -18^2 + 3

[Phy124]

can anyone pls help?

I believe it should be as 0.1057 is the probability of catching an undersized fish, which we don't want (they are ruled out) - although this is the first probability question I've attempted in 4 months, so could be wrong 

edit: Oh wait, never mind, I see what you're asking, you would round down to the lower whole number IIRC.

You would have . As you most likely can't catch 0.7803 of a fish, the answer would be 17.

help!

transform the graph y = 1/x^2

1. dilation of factor 2 from the x axis
2. dilation of factor 3 from the y axis
3. reflection in the x-axis
4. translation of 3 units in the positive direction of the y axis

i get y = 18/x^2 +6, but the answer says y= 18/x^2 +3....

edit: misread first quote

[Kanon]

Lets give this one another gooo





Stuck at this point, it's a m/c question so thats why I took Log10 because all the answers excluding one ae all Log10.  My question, why wouldn't we take Log3 so taht it cancels out with the exponential component of the eqn? 

[rife168]

Lets give this one another gooo





Stuck at this point, it's a m/c question so thats why I took Log10 because all the answers excluding one ae all Log10.  My question, why wouldn't we take Log3 so taht it cancels out with the exponential component of the eqn? 

From







Is that one of the choices?(I've tried to incorporate base 10) If not, would you be able to post what they are?

[Kanon]

Lets give this one another gooo





Stuck at this point, it's a m/c question so thats why I took Log10 because all the answers excluding one ae all Log10.  My question, why wouldn't we take Log3 so taht it cancels out with the exponential component of the eqn? 

From







Is that one of the choices?(I've tried to incorporate base 10) If not, would you be able to post what they are?

I wasn't aware you were able to rearrange two logs into an exponential w/ a log!
Mind = Blown!!11!1elevenhundredandone!!1
Bit confused with the last step though, how were you able to covnert log3 to log10?

The answers that look similar to yours are



Do you know where the -1 could have come from?

[rife168]

Lets give this one another gooo





Stuck at this point, it's a m/c question so thats why I took Log10 because all the answers excluding one ae all Log10.  My question, why wouldn't we take Log3 so taht it cancels out with the exponential component of the eqn? 

From







Is that one of the choices?(I've tried to incorporate base 10) If not, would you be able to post what they are?

I wasn't aware you were able to rearrange two logs into an exponential w/ a log!
Mind = Blown!!11!1elevenhundredandone!!1
Bit confused with the last step though, how were you able to covnert log3 to log10?

The answers that look similar to yours are



Do you know where the -1 could have come from?

It seems that they have simplified at my 3rd step by subtracting 1 from both sides and then dividing by 3. The -1 is the result from 'taking the 1 to the other side'.

The change of base rule states that      the base can be anything as long as it is the same for the numerator and denominator, in this case it is 10

What do you mean by:

I wasn't aware you were able to rearrange two logs into an exponential w/ a log!

?

Anyway, the second option you mentioned is equivalent to what I gave, so I would choose that as my answer.



edit: quoted wrong bit of text :S
Also, a derivation for the change of base rule can be found here.  I recommend looking at the derivation for this and also for most things that you come up against this year to get a more intuitive understanding of the topics.

[Insa]

Hey, can anyone lead me through this question? I've already substituted in the values to get two equations, but I don't know what the next step is.

If , and y = 19.6 when t = 2, and y = 19.02 when t = 5, find the value of the constants A and k. Give your answers correct to 2 decimal places.

Thank you

[oliverk94]

Hey, can anyone lead me through this question? I've already substituted in the values to get two equations, but I don't know what the next step is.

If , and y = 19.6 when t = 2, and y = 19.02 when t = 5, find the value of the constants A and k. Give your answers correct to 2 decimal places.

Thank you

Firstly you need sub in your values that you are given into the equation so that you have 2 equations, once you have them you can use simultaneous equations to solve for one of the unknowns (in this case, k) and then once you find that, you can sub that into one of the two equations (in this case equation [1]) to get find the other unknown (in this case, A).

19.6=Ae^-2k [1]
19.02=Ae^-5k [2]



[1]÷[2]

1.030494=e^3k
loge 1.030494 = 3k

loge 1.030494/3=k

k=0.01 [3]

Sub [3] into [1]

19.6=Ae^-2(0.01)
19.6=Ae^-0.02
19.6= 0.980199A
19.6/0.980199=A
A=20

So the equation will be y=20e^-0.01t