Preliminary mathematics marathon (1 Viewer)

[hscishard]

a= dv/dt = dv/dx * dx/dt
= dv/dx *v
= dv/dx * d/dv( 1/2 v^2) (cancel dv's)
= d/dx (1/2 v^2)

thats the correct way

That 2nd step is cool.
Third is even better.

 

[hscishard]

just checking, make sure u need radian mode on calculator

Makes sense to use radians.

1.98

 

[hscishard]

find the exact value of tan( 2 arccos(12/13))

Umm...1? Lol.

 

[random-1005]

Umm...1? Lol.

guessing you havent done inverse functions yet, dont wory, thats ones is a real easy one once u know the method

 

[random-1005]

Evaluate integral (0 to 2) of sqrt(4-x^2), use x=2sin(u)

 

[hscishard]

Haven't started Year 12 3unit yet. Only the 2unit parts

Oh yea, this is a prelim thread.

 

[random-1005]

Haven't started Year 12 3unit yet. Only the 2unit parts

Oh yea, this is a prelim thread.

k

lol i dnt pay attention as to what is prelim and what is not, i thought integration is technically hsc anyway??

 

[hscishard]

Yea integration is hsc. All your questions were hsc. Lol

 

[random-1005]

Yea integration is hsc. All your questions were hsc. Lol

thats cause prelim is too easy and boring

 

[random-1005]

solve absolute value ( x+7) = 5-2x

 

[hscishard]

-2/3. Knew it had some trick in it.

 

[random-1005]

integrate (4x^2 +3)^2 dx

 

[random-1005]

differentiate

y= ln ( (5x-1)/ ( x^3 -2x)^2) )

 

[hscishard]

(16/5)x^5 + 8x^3 + 9x

I don't think you can use sub method in that one..

 

[random-1005]

(16/5)x^5 + 8x^3 + 9x

I don't think you can use sub method in that one..

correct

 

[hscishard]

differentiate

y= ln ( (5x-1)/ ( x^3 -2x)^2) )

5(x^3-2x)^3 - 5(x-1)(9x^2-6) divided by (x^3-2x)^3

Theres an easier way, right?

 

[random-1005]

5(x^3-2x)^3 - 5(x-1)(9x^2-6) divided by (x^3-2x)^3

Theres an easier way, right?

using log laws

y= ln (5x-1) - ln ( 3x^2-2x)^2
y= ln(5x-1) -2ln(3x^2 -2x)

etc

im not sure if thats hsc, think it is lol

 

[hscishard]

Omg. I didn't see that

 

[random-1005]

if integral (1 to 5) of f (x) =10

find integral (1 to 5) {f(x) -5x^2}, dx

 

[hscishard]

if integral (1 to 5) of f (x) =10

find integral (1 to 5) {f(x) -5x^2}, dx

Umm...10-620/3. Cbf getting my calc out.