VCE Methods Question Thread!

[StupidProdigy]

Hey, I've forgotten how and when its appropriate to use the tangentline function on cas, could I use it on a question like: Find the equations of the tangent and normal for f(x)=2x^2-3x+4 at the point (3,13)? Thanks

[IndefatigableLover]

Hey, I've forgotten how and when its appropriate to use the tangentline function on cas, could I use it on a question like: Find the equations of the tangent and normal for f(x)=2x^2-3x+4 at the point (3,13)? Thanks

What you would do is: [Menu][4][9] and then input it in as:

tangentLine(2x^2-3x+4, x=3) and you'll find the tangent at that point which so happens to be y=9x-14

This is also the same for the normal as well

[StupidProdigy]

What you would do is: [Menu][4][9] and then input it in as:

tangentLine(2x^2-3x+4, x=3) and you'll find the tangent at that point which so happens to be y=9x-14

This is also the same for the normal as well

Thanks so much!! That'll save me heaps of time for my exam tomorrow!

[Zues]

Can someone help out on this. Possibly with the use of the addition law of probability formula

Using it I get

50= (36-x)+(20-x)-x

Where I get x to be two. However answer has +x at the end which gets x equals 6 which is right. Why is it plus and not minus? Thanks

[Zues]

Need help with the intersection part...

[Phy124]

It's because you're adding the left side (36 - x) and the right side (20 - x) but leaving out the middle (x).

Example with colours:



Piano = Red = 36
Guitar = Blue = 20
Piano & Guitar = Orange = x

Purple + Orange + Green = 50
Purple = Red - Orange
Green = Blue - Orange

(Red - Orange) + (Blue - Orange) + Orange = 50
(36 - x) + (20 - x) + x = 50
=> x = 6

[Zues]

Perfect. Thanks

[Zues]

Help please

[Zues]

and technically speaking, why is it not "B"

[Orb]

and technically speaking, why is it not "B"

Because amplitude is never negative.

You can think of amplitude as just the height of the peaks (so the distance from highest point and the lowest point) divided by two. You don't have a negative height, so it has to be A).

[Zues]

well i was thinking -(-a) = a so hence satisfying the "positive constants" restriction

[keltingmeith]

well i was thinking -(-a) = a so hence satisfying the "positive constants" restriction

But why would you take the negative, when they clearly say that a is ALREADY a positive constant?

[Zues]

why is this negative?

[DSubShell]

why is this negative?

Well, the question asks for . Since , then it logically follows that ... ie. The overall angle inside the Cos is in the 2nd Quadrant.

Cos(x) produces a negative result when "x" is in the 2nd quadrant (Just a fact you have to learn), so the overall answer must be negative!

[Zues]

Well, the question asks for . Since , then it logically follows that ... ie. The overall angle inside the Cos is in the 2nd Quadrant.

Cos(x) produces a negative result when "x" is in the 2nd quadrant (Just a fact you have to learn), so the overall answer must be negative!

ahhhh crap wow!! i totally thought that said sin (theta + ..) and not cos wow hahaha sorry for that, really stupid question!

But as i posted this, i was thinking of the rule where sin (90-theta) = cos theta (1st) and didnt read the question properly