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Here's a similar derivation, but makes explicit use of partial fractions:

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1220*5311=6479420^how to get ans 4947

1220*5311 = 6479420

6479420^x=4947

ln(6479420^x)=ln(4947)

xln(6479420)=ln(4947)

x=ln(4947)/ln(6479420)

x= ln(4947)/ln(6479420) = 0.5423654574711226

TRUE!   I just used 73107 kg......due to poor syntax in question..... it was still ridiculous... yah?

Not sure anyone (including YOU) really cares....Silly

ok

\(x=u^2+1\\ dx=2u\;\;du\\~\\ x-1=u^2\qquad u^2\ge0\;\;so\;\; x\ge1 \\ u=\pm\sqrt{x-1}\;\; \\ \text {Rather than having }x=u^2+1\\\text{I am going to have }u=+\sqrt{x-1 }\\ \text{Everything else is the same but I don't have to worry about the negative posibility}\\ \)

\(\displaystyle \int \frac{1}{2x\sqrt{x+1}}\;dx\\ =\displaystyle \int \frac{1}{2(u^2-1)u}\;2u\;\;du\\ =\displaystyle \int \frac{1}{u^2-1}\;\;du\\ =-\displaystyle \int \frac{1}{1-u^2}\;\;du\\ =-tanh^{-1}(u)+c\\ \qquad\text{equivalent for restricted values (from Wofram|alpha) }\\ \)

\(=\frac{1}{2} \left[ ln\frac{1-u}{1+u} \right]+c\\ =\frac{1}{2} \left[ ln\frac{1-u}{1+u} \right]+c\\ =\frac{1}{2} \left[ ln\frac{u-1}{u+1} \right]+c\\ \qquad \text{ Now c can be replaced with } logA\\ \qquad \text{Where A is a constant >0}\\ \qquad \text{I do not know why this condition was not given...}\\ =ln \left[ \frac{u-1}{u+1} \right]^{1/2}+lnA\\ =ln\left[A \left[ \frac{u-1}{u+1} \right]^{1/2}\right]\\ =ln\left[A\sqrt{ \left[ \frac{u-1}{u+1} \right]}\;\right]\\ \qquad u=\sqrt{x+1}\\ =ln\left[\;A\sqrt{ \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} }\;\;\;\right]\\ \)

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15a^2, 18ab

GCF  =   3a

06 and 60

15 and 51

24 and 42

33 and 33

The last one is obviously not realistic though because the father and the son can't both be 33, but I would still consider it a possible combination.

EP: You were too generous with the questioner in calculating the KE as:5,263,704 j. You took the mass of the Ocean Liner as 73,107 Kg, or ~73 tonnes. An Ocean Liner can in fact have a mass of between 50,000 - 100,000 metric tonnes. The questioner means: 7.3 x 10^7 Kg=73,000 metric tonnes!!. Therefore, the KE of the Ship would be:5,256,000,000 J.

(1/2)² = 1/4

cos (θ) / sin (θ)  =  cot(θ)

True. Should have thought fist before I posted, sorry.

tan x +cot x =8cos2x

sin(x) / cos (x)   + cos(x) / sin(x)  =  8cos (2x)

sin^2(x) + cos^2(x)  / [ sin(x)cos(x)]  =  8cos(2x)

1 / [sin(x)cos(x)] = 8cos(2x)

1/[ sin(2x)/2 ]   = 8cos(2x)

2 / sin(2x)  = 8sos(2x)

1/ sin(2x)  = 4cos(2x)

1/4  =  cos(2x)sin(2x)

1/4  =  (1/2)sin(4x)

2/4  =  sin(4x)

1/2 = sin(4x)

sin (4x)  = 1/2    when  4x = pi/6,   5pi/6,  13pi/6   and  17pi/6  

                            so      x = pi/24, 5pi/24, 13pi/24 and  17pi/24

Seriously?     What do you think would happen if you shot the liner head on with your little bullet?.....would it be pushed backwards or stopped?

KE = 1/2 m v^2    Bullet = 77.61 j   Liner = 5263704 j

sin (x)  = 0.8   = 4/5

Use the arcsin

arcsin (4/5)  = x  ≈  .927 = A   and  x  = pi – arcsin (4/5)  ≈ 2.214 = B

No it is not!

I do not think it is what you do!

I think you try your best to learn from all the answers you are given!

Is not that what everyone does?

4572 - 56788887 = -56784315

(1/25)^2

A 1 in twenty five chance of a homerun over 2 games = 0.0016 = 0.16%

It's on Khan Academy, and I've tried LOTS of times, and I don't think it's impossible since KA is a well funded and popular site, so if it was then someone would of already reported it, oh, and also I can't reveal answer because the reveal answer button is gone including the get a hint which just says "there is no hint" when I try clicking it.

I doubt very much that felrose has any idea what you have done Chris.

I doubt if he/she is even interested.

He/she is not using us to learn he is just using us to do his/her homework.