Mathematics Marathon (2 Viewers)

[lolokay]

Still confused about 'touches' and intersects.

what do you mean? touches = tangent, intersect = passes through (might include tangent as well)

 

[tommykins]

got it, read the question wrong

 

[Ali92l]

Since no ones posting questions =), ill keep the ball rolling, This is fairly simple...


A biased coin is tossed twice. The probability of getting AT LEAST one head is 5/9. Find the probability of getting 1 head

 

[bored of sc]

I can't do the above question since I... don't know how and haven't done topic.

So here's another:

Solve for x:

27^x = 3sqroot(3)

 

[lolokay]

A biased coin is tossed twice. The probability of getting AT LEAST one head is 5/9. Find the probability of getting 1 head

Spoiler

probability of getting tails both times is 1-5/9 = 4/9
the probability of getting tails in one toss is the square root of this, ie 2/3
so the probability of getting a head in 1 throw is 1/3, and the probability of getting heads twice is the square of this, ie. 1/9
so the probability of getting only 1 heads (and 1 tails) is P(at least one heads) - P(2 heads) = 5/9 - 1/9
= 4/9

"Solve cotA +4tanA =4cosecA for 0=<360"

Spoiler

cosA/sinA + 4sinA/cosA - 4/sinA = 0
multiplying through by sinAcosA
cos²A + 4sin²A - 4cosA = 0
cos²A + 4 - 4cos²A - 4cosA = 0
-3cos²A - 4cosA + 4 = 0
(-3cosA + 2)(cosA + 2)
cosA = -2 (no solution)
or cosA = 2/3
A = cos^-1[2/3]
= 48'11', 311'49'

fixed it

 

[ratcher0071]

I can't do the above question since I... don't know how and haven't done topic.

So here's another:

Solve for x:

27^x = 3sqroot(3)

sqrt 27 = 3sqrt3
27^1/2 = 3sqrt3
x= 1/2

 

[tommykins]

∫ 2e dx

Find the differential of xln(x).
Hence find ∫ ln(x) dx

 

[bored of sc]

Spoiler

probability of getting tails both times is 1-5/9 = 4/9
the probability of getting tails in one toss is the square root of this, ie 2/3
so the probability of getting a head in 1 throw is 1/3, and the probability of getting heads twice is the square of this, ie. 1/9
so the probability of getting only 1 heads (and 1 tails) is P(at least one heads) - P(2 heads) = 5/9 - 1/9
= 4/9

"Solve cotA +4tanA =4cosecA for 0=<360"

Spoiler

cosA/sinA + 4sinA/cosA - 4/sinA = 0
multiplying through by sinAcosA
cos²A + 4sin²A - 4cosA = 0
cos²A + 4 - 4cos²A - 4cosA = 0
-3cos²A - 4cosA + 4 = 0
(-3cosA + 2)(cosA + 2)
cosA = -2 (no solution)
or cosA = 2/3
A = cos^-1[2/3]
= 48'11'

You missed a solution in the second one. Cosine is also positive in the fourth quadrant and the domain is 0-360^o. So cos = 48'11' and 360-48'11'.

Anyone want to validate what I just said?

 

[lyounamu]

You missed a solution in the second one. Cosine is also positive in the fourth quadrant and the domain is 0-360^o. So cos = 48'11' and 360-48'11'.

Anyone want to validate what I just said?

He didn't because cosa =/= -2. The range is -1<=y<=1

 

[ratcher0071]

He didn't because cosa =/= -2. The range is -1<=y<=1

I think bored of sc is talking about the 2nd solution to Cos A =2/3

A = 48.19^o or A = 360^o - 48.19^o

A = 48.19^o or A = 311.81^o

 

[lyounamu]

I think bored of sc is talking about the 2nd solution to Cos A =2/3

A = 48.19^o or A = 360^o - 48.19^o

A = 48.19^o or A = 311.81^o

lol, ok. i thought he meant something else.

 

[Ali92l]

∫ 2e dx

Find the differential of xln(x).
Hence find

∫ 2e dx = 2ex + C

d xln(x) / dx = lnx + 1

∫ ln(x) dx = xlnx -1 + C



For what values of x does the following geometric series have a limiting sum

ln(x) + (ln(x))^2 + (ln(x))^3 .......

 

[bored of sc]

I think bored of sc is talking about the 2nd solution to Cos A =2/3

A = 48.19^o or A = 360^o - 48.19^o

A = 48.19^o or A = 311.81^o

YEAH! Thanks.

 

[zzdfa]

....(infinity)
.........Σ [n/(2^n)]
.......n=1

<!-- / icon and title --> <!-- message --> Uluru is a large rock on flat ground in Central Australia. Three tourists A, B, and C are observing Uluru from the ground. A is due north of Uluru, C is due east of Uluru, and B is on the line-of-sight from A to C and between them. The angles of elevation to the summit of Uluru from A, B, and C are 26 degrees, 28 degrees, and 30 degrees, respectively. Determine the bearing of B from Uluru with working.

this took me ages, i kept trying to use vectors and graphs (y=mx+c)
turns out it was just a simple sine rule question
however i think i still got it wrong, i got 4.1489 degrees, which doesn't sound right. here is my working.
http://img510.imageshack.us/my.php?image=mathsbz4.jpg

 

[zzdfa]

∫ 2e dx = 2ex + C

d xln(x) / dx = lnx + 1

∫ ln(x) dx = xlnx -1 + C



For what values of x does the following geometric series have a limiting sum

ln(x) + (ln(x))^2 + (ln(x))^3 .......

how do you do the integral sign?
and the answer should be xlnx +c right?

would the geometric series have a limitng sum for 0 < x < e?

 

[bored of sc]

how do you do the integral sign?

Just copy and paste it from their post.

 

[Ali92l]

how do you do the integral sign?
and the answer should be xlnx +c right?

would the geometric series have a limitng sum for 0 < x < e?

1/e < x < e

 

[M@ster P]

differentiate: (x^6 - x^3 + 6x + 3)^6

 

[clintmyster]

y' = 6(6x^5 -3x^2+6)(x^6 - x^3 + 6x + 3)^5
= 18(2x^5-x^2+2)(x^6 - x^3 + 6x + 3)^5

 

[M@ster P]

correct, i just made up that question lol