VCE Methods Question Thread!

[cutiepie30]

Hello everyone ,

for this question i was wondering how i would do it using x' and y' notation so i could states the sequence of transformations from one graph to the other.

Thanks 

[keltingmeith]

A tug-o-war team produces a tension in a rope described by the rule T=290(8t−0.5t2−1.4) units, where t is the number of seconds after commencing the pull.

1. Sketch a graph of T against t, stating the practical domain.
2. What is the greatest tension produced during a ‘heave’?

At the moment i am clueless and don't know where to start so it would be great if someone could help!

Okay, so for the first part, you'll need to brush up on your sketching skills. The easiest way I find to sketch a parabola is to find 3 parts - the x-intercepts, the y-intercepts, and the turning point. Once you've got all of those three done, it's just a matter of getting a line to go through all parts.

The second part, you want to know the most tension that the rope is going to experience. Hopefully this'll become clearer once you've sketched the graph. Let us know if you're confused by anything else!

Hello everyone ,

for this question i was wondering how i would do it using x' and y' notation so i could states the sequence of transformations from one graph to the other.

Thanks 

Dude, it hadn't even been a day since you asked, y'all need to chillax. We're not your personal tutors, remember that. For this question, all you need to do is figure out what you did to the equation to get from one side to the other. It might help if you separate things out like so:

Equation 1:


Equation 2:

[Mudasser.abb]

What happens if you dont write in pen for the methods exam?

[whys]

What happens if you dont write in pen for the methods exam?

Your writing may not scan as pencil may be too light for the scanner to pick up in some instances. To be safe, go with pen. It's too much of a risk to write in pencil, because if it doesn't scan properly, you may as well end up getting 0 (or close to it). I know there have been people who wrote in pencil (dark, such as 2B) but it doesn't hurt to be on the safe side and just use pen.

[cutiepie30]

Dude, it hadn't even been a day since you asked, y'all need to chillax. We're not your personal tutors, remember that. For this question, all you need to do is figure out what you did to the equation to get from one side to the other. It might help if you separate things out like so:

Equation 1:


Equation 2:

Thanks for your help keltingmeith


I dont know what to do after this like what does x' and y' equal still confused ??

[amanaazim]

can someone please explain ands love this question for me

Question: The equation of the line that passes through the point (5,9) is parallel to the line y=3x+7 is

how do you solve this on CAS (TI- nspire)

2=(2k-2)(3-k)


like can you tell me what do i have to put on the calculator

Mod edit: merged double post. In future please edit/modify your existing post rather than posting twice in a row

[TheEagle]

can someone please explain ands love this question for me

Question: The equation of the line that passes through the point (5,9) is parallel to the line y=3x+7 is

With these sorts of questions, I like to first refer to the original given equation. y=3x+7.    If a line was to be parallel to this, then the gradient (the steepness) needs to be the same because they will not touch/cross. So now our equation is: y= 3x+C
They have given us a pair of coordinates, (5,9), so we can use this to find C.   9=3*(5)+C  therefore C= -6
Therefore our equation is y=3x-6

[colline]

how do you solve this on CAS (TI- nspire)

2=(2k-2)(3-k)


like can you tell me what do i have to put on the calculator

Solve(2=(2k-2)(3-k),k)

[TheEagle]

how do you solve this on CAS (TI- nspire)

2=(2k-2)(3-k)


like can you tell me what do i have to put on the calculator

Menu, 3, 1,
Solve (2=(2k-2)(3-k),k)
the comma k means you are solving for that variable

[amanaazim]

Menu, 3, 1,
Solve (2=(2k-2)(3-k),k)
the comma k means you are solving for that variable

thank you so much. It worked

[amanaazim]

can someone help me solve this:

question: write the equation of the line that is perpendicular to the line 2y+x=2017 and passing through the point (-2,1)

basically i know that you make y the subject and put into y=mx+c form which i did like this

2y=-x+2017
y=-x/2+ 2017/2  (note that / means divide cause i don't know how to get that sign that you use in fractions, someone please let me know.)
AND for that answer i got m= -1/2.

so after this what do i do.

[Evolio]

Hello. 

Okay, so you found the gradient of the line, which is -1/2.
A perpendicular line is one that has a negative reciprocal of the original line (above line). That is, if m is the gradient, then the gradient of the perpendicular line is -1/m. So, this means that the gradient of the perpendicular line is -1/-1/2 which is equal to 2.
So, the perpendicular line has gradient 2 (m=2) , so it has equation y=2x+c. Then, you sub in the point (-2,1) where you substitute x=-2 and y=1 to get the value of c.
So: 1=2(-2)+c which is 1=-4+c. Add 4 to both sides. 1+4=c. Thus, c=5.
Then you have your equation y=2x+5

With these questions where you need to find the parallel line or the perpendicular line, you need to ONLY consider the gradient. The gradient is important. In a parallel line, the gradient is the same while for a perpendicular line, the gradient is a negative reciprocal where if m is the gradient of the original line, the gradient of the perpendicular line would be =-1/m.

[amanaazim]

Thank you so much.I actually understood the way you worded the solution for my problem. Thank you so much for helping on this and my previous posts

[amanaazim]

can someone solve this please.

Find the equation of the line perpendicular to the line that has an equation 2x-4y+3=0
and passes through (-1,2)

[TheEagle]

can someone solve this please.

Find the equation of the line perpendicular to the line that has an equation 2x-4y+3=0
and passes through (-1,2)

This follows the same principal but you need to first set the equation in the form of y=mx+c
therefore, -4y=-2x-3
y=-2/-4 x -3/-4
y=1/2x + 3/4

m of the perpendicular is -1/m
-1/(1/2) = -1*2 = -2

Perpendicular equation : y=-2x+c   (-1,2)

2= -2*(-1)+c.  therefore c= 0


therefore perpendicular equation is y=-2x