Mathematics Marathon (4 Viewers)

[gurmies]

a^2 = b^2 + c^2 - 2bcCosA

25 = 75 + x^2 - (2)(5root3)(x)(root3/2)

25 = 75 + x^2 - 15x

x^2 - 15x + 50 = 0

(x-10)(x-5) = 0

x = 10 or/ x = 5

Find the derivative of root(sin2x)

 

[shaon0]

a^2 = b^2 + c^2 - 2bcCosA

25 = 75 + x^2 - (2)(5root3)(x)(root3/2)

25 = 75 + x^2 - 15x

x^2 - 15x + 50 = 0

(x-10)(x-5) = 0

x = 10 or/ x = 5

Find the derivative of root(sin2x)

f(x)=sqrt(sin2x)
Let u=sin2x
f'(x)=cos(2x)/sqrt(sin2x)

Derive cot(arcsin(x))=sqrt(1-x^2)/x
Sorry, don't know if this is 2unit. but it's fairly short.

 

[jet]

Let theta = arcsin(x)

Hence sin(theta) = x/1

This leads to a right triangle with one side x, the hypotenuse 1 and the other side sqrt(1-x^2)

hence, cot(theta) = 1/tan(theta)
=sqrt(1-x^2)/x

Thats definitely not 2 unit.

How about:
The curve x^2/5 + y^2/17 = 10 in the first quadrant is rotated through the x-axis.
Find the volume of the solid of revolution created by this rotation.

 

[shaon0]

Let theta = arcsin(x)

Hence sin(theta) = x/1

This leads to a right triangle with one side x, the hypotenuse 1 and the other side sqrt(1-x^2)

hence, cot(theta) = 1/tan(theta)
=sqrt(1-x^2)/x

Thats definitely not 2 unit.

Sorry about that.
My solution was:
Let u=arcsin(x)
and, cot(u)=cos(u)/sin(u).
cos(arcsin(x))/x=cot(arcsin(x))
Since, x=sin(u)
x^2=1-(cos(u))^2
cos(u)=sqrt(1-x^2)
Hence, cot(arcsin(x))=sqrt(1-x^2)/x

How about:
The curve x^2/5 + y^2/17 = 10 in the first quadrant is rotated through the x-axis.
Find the volume of the solid of revolution created by this rotation.

17x^2+5y^2=850
5y^2=850-17x^2
y^2=170-(17/5)x^2

V=pi S [limits: sqrt(5) to 0] y^2 dx
= pi S (170-(17/5)x^2) dx
= pi (70sqrt(5)-(17/15)sqrt(5)^3)
= 70(pi)sqrt(5)-(17pi/3)sqrt(5)

 

[jet]

17x^2+5y^2=850
5y^2=850-17x^2
y^2=170-(17/5)x^2

V=pi S [limits: sqrt(5) to 0] y^2 dx
= pi S (170-(17/5)x^2) dx
= pi (70sqrt(5)-(17/15)sqrt(5)^3)
= 70(pi)sqrt(5)-(17pi/3)sqrt(5)

You got the limits wrong. Its not the standard equation for an ellipse

Think like a 2 unit person.

 

[shaon0]

You got the limits wrong. Its not the standard equation for an ellipse

Think like a 2 unit person.

17x^2+5y^2=850
5y^2=850-17x^2
y^2=170-(17/5)x^2

V=pi S [limits: sqrt(50) to 0] y^2 dx
= pi S (170-(17/5)x^2) dx
= pi (70sqrt(50)-(17/15)sqrt(50)^3)
= 350(pi)sqrt(2)-(850(pi)/3)sqrt(2) <-------- This part might be wrong because i cbf using a calculator.

 

[Kaatie]

cube root(27x^27)

 

[gurmies]

cube root(27x^27)

3x^9

Differentiate tan x°

 

[shaon0]

3x^9

Differentiate tan x°

By 1st principles?
d/dx(tan(x))= lim (h->0)(tan(x+h)-tan(x))/h
= lim (h->0)((tan(x)+tan(h)/1-tan(x)tan(h)-tan(x))/h
= lim (h->0)(tan(h)+(tan(x))^2*tan(h))/h
= lim (x->0)(tan(h)/h).lim (x->0)(1+(tan(x))^2)
= pi/180° (sec(x))^2

 

[Kaatie]

no question again?

okay than heres an odd one:
The face of a 50c coin is a regular dodecagon. Show that its volume is approx 95.5% i.e. 3/pie of that of a circular coin of the same diameter and thickness

 

[Kaatie]

if no one knows how to do it feel free to ask a new question

 

[jet]

Okay.
A dodecagon will have 12 isosceles triangles, with the two equal sides r and the other side b, and angles 30, 75 and 75.

By constructing the perpendicular height of length h from the base to the vertex at the centre, we have:
sin15 = (b/2)/r = b/2r
Hence, b = 2rsin15

Also, cos15 = h/r
h = rcos15

Now, the area of each triangle, a = 0.5 x 2rsin15 x rcos15
=r^2 sin15 cos15

Hence, the area of the dodecagon is A = 12r^2 sin15 cos 15, and the volume, V1 = 12r^2Tsin15 cos15 (thickness T)

For a circular coin, the volume is V = πr^2 T

Hence, the % Volume = V1/V x 100
= (12r^2 Tsin15 cos15)/(πr^2 T) x 100
=(1200sin15 cos15)/(π)
=95.493% as required.

Solve for x:
3^(2x + 1) - 180(3^x) + 729 = 0

 

[bored.of.u]

Okay.
A dodecagon will have 12 isosceles triangles, with the two equal sides r and the other side b, and angles 30, 75 and 75.

By constructing the perpendicular height of length h from the base to the vertex at the centre, we have:
sin15 = (b/2)/r = b/2r
Hence, b = 2rsin15

Also, cos15 = h/r
h = rcos15

Now, the area of each triangle, a = 0.5 x 2rsin15 x rcos15
=r^2 sin15 cos15

Hence, the area of the dodecagon is A = 12r^2 sin15 cos 15, and the volume, V1 = 12r^2Tsin15 cos15 (thickness T)

For a circular coin, the volume is V = πr^2 T

Hence, the % Volume = V1/V x 100
= (12r^2 Tsin15 cos15)/(πr^2 T) x 100
=(1200sin15 cos15)/(π)
=95.493% as required.

Solve for x:
3^(2x + 1) - 180(3^x) + 729 = 0

here my solution:

3(3^2x) -180(3^x) +729 = 0
3(3^x)^2 - 180(3^x) + 729 = 0
so let, u = (3^x)
.: 3 u^2 - 180u + 729 =0
3[(u^2) -60u + 243] = 0
then using the quadratic formulae
im gonna speed up the process a little =D
u=(60+/- SQROOT 3600-972)/2
u=(60+/- SQROOT 2628)/2
u=30+/- SQROOT 657
so evaluating to 2dp
u = 55.63 and 4.36

am i right??

n heres my question:

solve for x:

tan^4 x - tan^3 x - 3tan^2 x + 3tan x = 0 for 0 ≤ x ≤ 360

 

[gurmies]

here my solution:

3(3^2x) -180(3^x) +729 = 0
3(3^x)^2 - 180(3^x) + 729 = 0
so let, u = (3^x)
.: 3 u^2 - 180u + 729 =0
3[(u^2) -60u + 243] = 0
then using the quadratic formulae
im gonna speed up the process a little =D
u=(60+/- SQROOT 3600-972)/2
u=(60+/- SQROOT 2628)/2
u=30+/- SQROOT 657
so evaluating to 2dp
u = 55.63 and 4.36

am i right??

n heres my question:

solve for x:

tan^4 x - tan^3 x - 3tan^2 x + 3tan x = 0 for 0 ≤ x ≤ 360

You are correct but you would not get full marks, as the question said solve for x. You would have to proceed to use logarithms to solve 3^x = 55.63 and 4.36.

For your question:

tan^4(x) - tan^3(x) - 3tan^2(x) + 3tan(x) = 0

tan^3(x)[tan(x) - 1] -3tan(x)[tan(x) - 1] = 0

[tan(x) - 1][tan^3(x) - 3tan(x)] = 0

tan(x)[tan(x) - 1][tan^2(x) - 3] = 0

tan(x)[tan(x) - 1][tan(x) - √3][tan(x) + √3] = 0

tan(x) = 0

tan (x) = 1

tan(x) = √3

tan(x) = -√3

x = 0, 180, 360, 45, 225, 60, 240, 120, 300

Prove that the parabolas y = 2x^2 - 6x + 7 and y = x^2 - 2x + 3 touch eachother and find the co-ordinates of the point of contact (Fitzpatrick 2 Unit)

 

[jet]

Whoops, sorry. The question is meant to be
3^(2x + 2) - 108(3^x) + 729 = 0

 

[bored of sc]

Solve for x:
3^(2x + 2) - 180(3^x) + 729 = 0

3^2x+2-180*3^x+729 = 0
3^2x*9-180*3^x+729 = 0
9(3^x)²-180(3^x)+729 = 0
Let 3^x = m
9m²-180m+729 = 0
m²-20m+81 = 0
m = [20+sqrt(20²-4*1*81)]/2
= [20+sqrt(76)]/2
= [20+2*sqrt(19)]/2
= 10+sqrt(19)
3^x = m
3^x = 10+sqrt(19)
x = log₃[10+sqrt(19)]

Not sure since I haven't done the logarithms unit yet.

 

[Nevermore]

3^2x+2-180*3^x+729 = 0
3^2x*9-180*3^x+729 = 0
9(3^x)²-180(3^x)+729 = 0
Let 3^x = m
9m²-180m+729 = 0
m²-20m+81 = 0
m = [20+sqrt(20²-4*1*81)]/2
= [20+sqrt(76)]/2
= [20+2*sqrt(19)]/2
= 10+sqrt(19)
3^x = m
3^x = 10+sqrt(19)
x = log₃[10+sqrt(19)]

Not sure since I haven't done the logarithms unit yet.

starting from 3^x = 10+sqrt(19)
log[3^x] = log[10+sqrt(19)]
x log 3 = log[10+sqrt(19)]

x = (log 3 / log[10+sqrt(19)])

 

[bored of sc]

New question: If there are 100,000 people doing the HSC in the year 2100 and you achieve a UAI of 95 which places you in the top 8% of those completing the HSC, how many people dropped out in year 10? I hope that makes sense and isn't flawed.

 

[Nevermore]

New question: If there are 100,000 people doing the HSC in the year 2100 and you achieve a UAI of 95 which places you in the top 8% of those completing the HSC, how many people dropped out in year 10? I hope that makes sense and isn't flawed.

from the problem, this equation is formed


5000
-------------- = 0.08
100000 - x

after solving for x

the answer is 37,500

 

[Nevermore]

Now here is my question

In a class of 28 students, students must choose to study at least one of the following languages - French, German.

22 study french and 18 study german, what is the probability of choosing a student who studies both languages?