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lolokay

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tommykins said:
Still confused about 'touches' and intersects.
what do you mean? touches = tangent, intersect = passes through (might include tangent as well)
 

Ali92l

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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
 
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bored of sc

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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:

27x = 3sqroot(3)
 

lolokay

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Ali92l said:
A biased coin is tossed twice. The probability of getting AT LEAST one head is 5/9. Find the probability of getting 1 head
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"
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
 
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ratcher0071

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bored of sc said:
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:

27x = 3sqroot(3)
sqrt 27 = 3sqrt3
271/2 = 3sqrt3
x= 1/2
 

tommykins

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∫ 2e dx

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

bored of sc

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lolokay said:
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"
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 0-360o. So cos = 48'11' and 360-48'11'.

Anyone want to validate what I just said?
 

lyounamu

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bored of sc said:
You missed a solution in the second one. Cosine is also positive in the fourth quadrant and the domain is 0-360o. 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

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lyounamu said:
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 = 48.19o or A = 360o - 48.19o

A = 48.19o or A = 311.81o
 

lyounamu

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ratcher0071 said:
I think bored of sc is talking about the 2nd solution to Cos A =2/3

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

A = 48.19o or A = 311.81o
lol, ok. i thought he meant something else.
 

Ali92l

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tommykins said:
∫ 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

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ratcher0071 said:
I think bored of sc is talking about the 2nd solution to Cos A =2/3

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

A = 48.19o or A = 311.81o
YEAH! Thanks.
 

zzdfa

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untouchablecuz said:
....(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

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Ali92l said:
∫ 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?
 

Ali92l

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zzdfa said:
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
 

clintmyster

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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
 

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