Mathematics Question Thread

[jamonwindeyer]

Oh I just realised that I misread part a) solution that Rui answered. When checking I kept typing (3/5)^3*3 instead of what it should have been. However didn't I do part b) wrong as (3/5)^2 * (2/5)^3 + (3/5)^3 = 747/3125 instead of it being my answer of 81/125?

No Part B looks okay in the version you posted! You've got:



That's correct! The 54/125 comes from Part A

[Fahim486]

Hi everyone,

I was just wondering anytime we do working outs for integration, do we have to use the integral sign in a step, like for example if a question asks to integrate d^2y/dx^2 = x^2+3, do we have to write dy/dx = (integral sign) x^2+3 and then actually integrate or can we skip that step and integrate straight away. I wanted to ask this question because my tutor says you have to so that the HSC markers don't get confused but my school maths teacher has never said anything like this or taught to do this.

Thanks!

[jamonwindeyer]

Hi everyone,

I was just wondering anytime we do working outs for integration, do we have to use the integral sign in a step, like for example if a question asks to integrate d^2y/dx^2 = x^2+3, do we have to write dy/dx = (integral sign) x^2+3 and then actually integrate or can we skip that step and integrate straight away. I wanted to ask this question because my tutor says you have to so that the HSC markers don't get confused but my school maths teacher has never said anything like this or taught to do this.

Thanks!

Hey! You can definitely integrate straight away if you need to - The integral sign is not a bad thing but it doesn't overly help what you are doing, in terms of clarity at least - It is pretty clear you've integrated based on the LHS of your expression

I always just integrated, if that puts you at ease

[bananna]

Hi!

I'm revising yr 11 work because I forgot a lot of content

Can someone pls help me with these q's

1) what is the solution to the equation 2cosβ = -√3 for 0° ≤ β ≤ 360° ?

2) what is the solution to the eqn cos(θ/2 + 20°) for  0° ≤ θ ≤ 90° ?


Thank you

[kiwiberry]

Hi!

I'm revising yr 11 work because I forgot a lot of content

Can someone pls help me with these q's

1) what is the solution to the equation 2cosβ = -√3 for 0° ≤ β ≤ 360° ?

2) what is the solution to the eqn cos(θ/2 + 20°) for  0° ≤ θ ≤ 90° ?


Thank you

For 1), cosβ=-√3/2
Remember that cos is negative in the second and third quadrants
So β=150°, 210°!

For 2), did you forget to finish typing the equation? But use the same method as the first, solve for θ/2 + 20° and then rearrange to find θ

[jamonwindeyer]

Hi!

I'm revising yr 11 work because I forgot a lot of content

Can someone pls help me with these q's

1) what is the solution to the equation 2cosβ = -√3 for 0° ≤ β ≤ 360° ?

2) what is the solution to the eqn cos(θ/2 + 20°) for  0° ≤ θ ≤ 90° ?


Thank you

Just to add to kiwi berry's answer, remember that the exact ratio results are on your reference sheet if you've forgotten them (the \(\sin{\theta}=\frac{1}{2}\implies\theta=30^\circ\) stuff!)

I'd be happy to show the full working for that second example if you provide the full question - Just to show you how to tackle that sort of question that needs rearranging

[smile123]

Equation of a line passing through (−5,7) and having infinite slope will be:
PLEASE HELP

[RuiAce]

Equation of a line passing through (−5,7) and having infinite slope will be:
PLEASE HELP

A line with infinite slope is just a vertical line.

All vertical lines are of the form x = something

So since the point (-5,7) passes through it, i.e. a point where x=-5, the line is also x=-5

[katnisschung]

find the area between the two curves: y=-3x+4 & y=x^3 and the x-axis

i drew the graphs but i don't know how to find the point of intersection
using simultaneous equations...

i got x^3+3x-4=0... do 2 unit need to know how to simplify an equation with a cube power specific to this case?

[katnisschung]

stuck on another question
find the area between  y=x^2+2x-8
and y=2x+1

again i sketched the graphs,

found their point of intersection to be 3 and -3 using simultaneous equations
but i'm having trouble finding what i need to integrate??

the areas are not bounded by the x-axis, in fact they fall below... so
do i need to split it up becos some of the areas below would be negative?

confused...essentially pls show me the working out thanks 

[kiwiberry]

find the area between the two curves: y=-3x+4 & y=x^3 and the x-axis

i drew the graphs but i don't know how to find the point of intersection
using simultaneous equations...

i got x^3+3x-4=0... do 2 unit need to know how to simplify an equation with a cube power specific to this case?

We can use the factor theorem and test some numbers to try and find a factor!
If P(x)=x³+3x-4, P(a)=0 when x-a is a factor
Subbing x=1, P(1)=1+3-4=0 so (x-1) is a factor
From here you can long divide to find the other factor and you'll get (x-1)(x²+x+4)=0
So the POI is at x=1

[RuiAce]

We can use the factor theorem and test some numbers to try and find a factor!
If P(x)=x³+3x-4, P(a)=0 when x-a is a factor
Subbing x=1, P(1)=1+3-4=0 so (x-1) is a factor
From here you can long divide to find the other factor and you'll get (x-1)(x²+x+4)=0
So the POI is at x=1

Problem. This is the 2U section, where polynomials are not taught. Therefore

find the area between the two curves: y=-3x+4 & y=x^3 and the x-axis

i drew the graphs but i don't know how to find the point of intersection
using simultaneous equations...

i got x^3+3x-4=0... do 2 unit need to know how to simplify an equation with a cube power specific to this case?

No, you are not. At least not for weird ones.

[kiwiberry]

Problem. This is the 2U section, where polynomials are not taught.

Oh oops sorry, my bad 😅

[itswags98]

Hey im trying to revise. Do you think you could help me out there with an explanation and working out of each of them? Thank you.

[RuiAce]

Hey im trying to revise. Do you think you could help me out there with an explanation and working out of each of them? Thank you.

__________________________________



__________________________________







Ok, I think E is just a dud.