Mathematics Marathon HSC 09 (6 Viewers)

[s2Vicki]

Question:

Solve 3-2[cos(^2)x]-2=0 for 0 degress<= x <= 260 degrees

 

[xFusion]

Question:

Solve 3-2[cos(^2)x]-2=0 for 0 degress<= x <= 260 degrees

whoops my bad. XD
(cosx)^2=1/2
x=45, 315. (assuming you ment 0<=x<=360)

Question thats still not done:Water flows from a full tank at the rate given by dh/dt= -1.5 sqrt(t)
where h is the depth of the water in meters at any time t minutes. Height of tank is 20m.
a)find the equation for h in terms of t
b)how long does it take for the tank to empty?

 

[AlexJB]

I think you wrote the question wrong again because you cant root a negative..in 2u

3-2[cos(^2)x]-2=0
1 - 2[cos(^2)x] = 0
2[cos(^2)x] = 1
cos(^2)x = 1/2
cosx = +- sqr.root (1/2)

Solve from there.

 

[s2Vicki]

whoops my bad. XD
(cosx)^2=1/2
x=45, 315. (assuming you ment 0<=x<=360)

whoops, yeah it was meant to be 360.

 

[s2Vicki]

actually i wrote the question wrong

it's actually 3-2(cos(x^2))x - 3sinx =0

and i got it mixed up with 2sin(^2)x + cos x -2 = 0

mind doing both of them? both are between 0 and 360 degrees

 

[AlexJB]

whoops my bad. XD
(cosx)^2=1/2
x=45, 315. (assuming you ment 0<=x<=360)

Question thats still not done:Water flows from a full tank at the rate given by dh/dt= -1.5 sqrt(t)
where h is the depth of the water in meters at any time t minutes.
a)find the equation for h in terms of t
b)how long does it take for the tank to empty?

More information needed?
h = -t sqrt(t) + c
Don't you need to be supplied the initial height if this is to work?

 

[AlexJB]

actually i wrote the question wrong

it's actually 3-2(cos(x^2))x - 3sinx =0

and i got it mixed up with 2sin(^2)x + cos x -2 = 0

mind doing both of them? both are between 0 and 360 degrees

Can you type the first one up on this? Online LaTeX Equation Editor

Second:
2sin(^2)x + cos x -2 = 0
2(1 - cos(^2)x) + cosx - 2 = 0
2 - 2cos(^2)x + cosx - 2 = 0
2 cos(^2)x - cosx = 0
cosx(2cosx - 1) = 0
cosx = 0 or cosx = 1/2
Solve for x

 

[s2Vicki]

Can you type the first one up on this? Online LaTeX Equation Editor

Second:
2sin(^2)x + cos x -2 = 0
2(1 - cos(^2)x) + cosx - 2 = 0
2 - 2cos(^2)x + cosx - 2 = 0
2 cos(^2)x - cosx = 0
cosx(2cosx - 1) = 0
cosx = 0 or cosx = 1/2
Solve for x

<a href="http://www.codecogs.com/eqnedit.php?latex=3-2cos^{2}x-3sinx = 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?3-2cos^{2}x-3sinx = 0" title="3-2cos^{2}x-3sinx = 0" /></a>

 

[xFusion]

actually i wrote the question wrong

it's actually 3-2(cos(x^2))x - 3sinx =0

and i got it mixed up with 2sin(^2)x + cos x -2 = 0

mind doing both of them? both are between 0 and 360 degrees

 

[Trebla]

Prove that a triangle with sides a, b and c has an obtuse angle if a² + b² < c² (hint: consider the cosine rule)

 

[xFusion]

More information needed?
h = -t sqrt(t) + c
Don't you need to be supplied the initial height if this is to work?

Oh yeah lol i forgot to say height of tank is 20m

 

[untouchablecuz]

Prove that a triangle with sides a, b and c has an obtuse angle if a² + b² < c² (hint: consider the cosine rule)

c² = a²+b²-2abcos(x)

where x is obtuse

x can be written in the form 90+y, where y is acute

i.e. cos(x) = cos(90+y) = sin(90-(90+y)) = sin(-y) = -sin(y)

hence, c² = a²+b²+2absin(y) > a²+b², since sin(y)>0

.'. a²+b² < c²

 

[ninetypercent]

who memorises the standard integrals that the BOS will give to us?

 

[boxhunter91]

I dont memorise. If you dont know them by the time you walk into the HSC exam i'd be worried :/

 

[khorne]

Here's a big cup of harden the hell up:

Find the volume of the ellipsoidal solid given by 2x^2 + 4y^2 + z^2 = 16

First ever maths troll

 

[untouchablecuz]

doesn't that require triple integrals

 

[khorne]

Damn right, bitches

 

[untouchablecuz]

 

[khorne]

 

[Trebla]

c² = a²+b²-2abcos(x)

where x is obtuse

x can be written in the form 90+y, where y is acute

i.e. cos(x) = cos(90+y) = sin(90-(90+y)) = sin(-y) = -sin(y)

hence, c² = a²+b²+2absin(y) > a²+b², since sin(y)>0

.'. a²+b² < c²

Perhaps a much neater approach would be:
cos θ = (a² + b² - c²) / 2ab
If θ is obtuse then cos θ < 0
=> a² + b² - c² < 0 (a, b are positive)
=> a² + b² < c²