3 Unit Maths HSC Exam Revision (2 Viewers)

[random-1006]

isn't it 0?

yes, but why arent you saying it with confidence!

 

[Trebla]

...because you're differentiating a constant

What about this?

 

[random-1006]

...because you're differentiating a constant

What about this?

let f(t) = e^(t^2)
= d/dx [integral f(t) dt between limits]
= d/dx [ F(x^2) - F(0) ]
= 2x F ' (x^2) - 0 * F ' (0)

but F ' (t) = f(t)

so answer = 2xe^(x^4)

but thats not 3 unit or 4 unit

 

[cutemouse]

...because you're differentiating a constant

What about this?

2xe^(x^4)

This is from the first fundamental theorem of calculus... I think.

 

[smik11]

LOL fitzpatrick, projectile motion chapter, question 17, i believe there are answers lying around on BoS somewhere

omg i want it!

 

[random-1006]

 

[hscishard]

-1 < x < 1, x<-1
Would it be correct to just put x<1. Nvm
How long should it take to get that?

 

[random-1006]

-1 < x < 1, x<-1
Would it be correct to just put x<1. Nvm
How long should it take to get that?

i dont know lol. as long as it takes.

it would be worth 3 marks in an exam i think

 

[random-1006]

 

[hscishard]

It's harder than the other one.
(1-0.5^18)/8 -549 = -548.875

 

[random-1006]

It's harder than the other one.
(1-0.5^18)/8 -549 = -548.875

something like that was asked on a past catholic trial, still easy, justing splitting into an AP and a GP

 

[bouncing]

hey guys lol this is such a stupid question but how do you do auxilliary/subsidiary angles again?
this is a sub/aux question right?

@= alpha (cbf using latex)

4cos@+3sin@=1

find @

 

[random-1006]

hey guys lol this is such a stupid question but how do you do auxilliary/subsidiary angles again?
this is a sub/aux question right?

@= alpha (cbf using latex)

4cos@+3sin@=1

find @

Let 4cos@ + 3sin@= Acos(@ + a)
therefore 4cos@ + 3 sin@= Acosacos@ - Asinasin@

equating coeffiencents

Acosa= 4 --- 1
Asina= -3 --- 2

sqaure and add 1 and 2
A^2= 25, A= 5, take positive case

now cos is positive and sin is negative in the fourth quadrant.

so we have cosa = 4/5 , related angle = 36.9 degrees, as we are in fourth quadrant, 360- 36.9= 323.1

so we have 5cos(@ + 323.1) = 1
cos(@ + 323.1) = 1/5 etc

 

[Hayzazz]

Shouldn't 4cos@ + 3sin@ = Acos(@ - a) ? Not Acos(@ + a)

EDIT:
Oh wait nvm it doesn't make a difference :\

 

[random-1006]

Shouldn't 4cos@ + 3sin@ = Acos(@ - a) ? Not Acos(@ + a)

im fairly sure from memory you can pick either. ie instead of getting 5cos(@+330) , you will get 5cos(@-30) ie 360 has been subtracted

but yes the cos(@-a) would look better, i wasnt think bout that

 

[random-1006]

 

[x jiim]

sorry about the awkward spacing, this is my first time using latex.

Solve

 

[mysticwater]

Not sure for the above question but i think you use logs and u'll get a ans with d.p?

The gradient of a curve is given by y'= . the curbe passes through points (0,2)

Find the question of the curve. How would u answ this?

 

[randomnessss]

QUESTION:

 

[hscishard]

sorry about the awkward spacing, this is my first time using latex.

Solve

(-ln3 - ln2) / (ln2 - ln3)
Beaten rofl.