VCE Specialist 3/4 Question Thread!

[barydos]

It's me again with another question!


I can't seem to get part d).. Not sure what it means by "circular arc", I guess.

Anyway, answers for part a-d are in spoiler tags below:

Spoiler

[pi]

A circular arc as in AB would be a tangent to a circumference at A of which AP would be an arc of, if that makes sense Try finding a point O such that AO and PO are radii (ie O is center of circle).

[barydos]

A circular arc as in AB would be a tangent to a circumference at A of which AP would be an arc of, if that makes sense Try finding a point O such that AO and PO are radii (ie O is center of circle).

Just curious, which part of the question lets us know that the 'circle' is that way and not in the other direction (e.g. forming a semi-circle through the points A, P and B or similar)?

[pi]

Sorry, I think I misinterpreted the question too, reading it as A to P instead of A to B through P. I think you're correct in thinking it's a semicircle through all three points.

[barydos]

Sorry, I think I misinterpreted the question too, reading it as A to P instead of A to B through P. I think you're correct in thinking it's a semicircle through all three points.

If that were the case, would it make sense to just do ? Doing that calculation gives me 15.7, while the answer says 14.18. I'm going to try doing A to P in a circle but PB directly.

Edit: Well, that obviously doesn't work. :/  I'm going to sleep, hopefully someone comes up with something tomorrow

[Holmes]

How to graph this in a CAS calculator? I'm not sure how to do it in terms of y= ..... just by using the cas to solve for y, rather than doing it by hand:

(x^2/6) + (y^2/9) = 1

GL[/gOlow]W! (just testing the glow button to see how it works )

[polar]

How to graph this in a CAS calculator? I'm not sure how to do it in terms of y= ..... just by using the cas to solve for y, rather than doing it by hand:

(x^2/6) + (y^2/9) = 1

GL[/gOlow]W! (just testing the glow button to see how it works )

which calculator do you use?
TI: Guide to Using the TI-Nspire for SPECIALIST – The intricate and tightly packed (towards the bottom of the post)
ClassPad: go to conics - type in your equation like you write it up there (you don't really need the brackets though) and press the hyperbola icon at the top

[Holmes]

which calculator do you use?

tinspire

I went to that link, and I understand how to do it. Just a conceptual thing though, when I write the 'zeroes' at the start, what does it mean for the sketching of the graph? "Step 1: Enter in the graph bar zeros(equation, dependent variable)"

[b^3]

tinspire

I went to that link, and I understand how to do it. Just a conceptual thing though, when I write the 'zeroes' at the start, what does it mean for the sketching of the graph? "Step 1: Enter in the graph bar zeros(equation, dependent variable)"

What is basically means is that we try to put all the terms onto one side so that we end up with the zero on the other side, then that is what we put into the 'equation' part, the dependent variable is the variable that will be reliant on another variable, normally we have x being the independent variable and y beind the dependent variable.

Also something I will point out aswell, if you have one of the newer versions of the OS/newer version of they calc, they've made it easier to graph ellipses and hyperbolas and such. When on the entry bar line click [ctrl] [Menu] [1] [2] and pick the appropriate template.

[Homer]

not sure how to approach dis?

[availn]

not sure how to approach dis?

I'm assuming that those are quarter circles inside that square? The area of one of the light slivers is equal to the area of the square, minus the area of one of the quarter circles.

1² - 0.25pi*1²
= 1 - pi/4

As we now know the area of one light sliver, just subtract that area from one of the quarter circles.

0.25pi*1² - (1 - pi/4)
= pi/4 - 1 + pi/4
= pi/2 - 1 units²

[sin0001]

Having trouble with this one:
The series Sn is given by Sn = 2n^2 + 14n. The nth term, tn, is equal to:
A)4n+12
B)4n-12
C)2n+7
D)2n-7
E)2n+14

Another one:
The radius lenth of the cicumference of a triangle with side lengths 9 cm, 12 cm and 15 cm is:
A)4,5
B)6
C)7.5
D)10
E)12

[polar]

to quote some kid on this forum

first, note that

[dim_sim]

I have some vector questions!
If |a| = |b|, simplify (a+b)^2
The answer from the textbook is 2a^2 (1+cos)
And, if a=2i-4j+k and b=-4i+j+2k, find a unit vector perpendicular to both a and b.
Thanks

[polar]

I have some vector questions!
If |a| = |b|, simplify (a+b)^2
The answer from the textbook is 2a^2 (1+cos)

the book had (a+b)^2? or was it |a+b|^2? note that


as for your second question, the only method I can think of that you can use without crossing (lol) into methods not in the course is:
let this third vector be
from three simultaneous equations