Alan

There is another way of approaching this (you have to think in terms of factorizing 486 =  2*3⁵ and 1944 = 2³3⁵ for this):

Let a = log_n(486√2)  ...(1)

Then it's also true that a = log_2n(1944)   ...(2)

It's useful to write (1) as   a = log_n(2*3⁵√2) = log_n(3⁵2^3/2)   so that  n^a = 3⁵2^3/2  ...(3)

Similarly, with (2) we have  a = log_2n(2³3⁵) so that (2n)^a = 2³3⁵   or 2^an^a = 2³3⁵  ...(4)

Substituting from (3) into (4) we get  2^a3⁵2^3/2 = 2³3⁵ from which  2^a = 2^3/2 so that a = 3/2.

Putting this back into (3) we have n^3/2 = 3⁵2^3/2 so n = (3⁵2^3/2)^2/3  or

n = 2(3⁵)^2/3 = 2*3^10/3 = 2*(3⁹3¹)^1/3

n = 2*3³*3^1/3

Finally:  n = 54*3^1/3

Just enter this in the site calculator here (be sure to put it into radian mode first - there is a button you can select near the bottom left of the calculator).

NB You could also simply enter cos(4/7)² and get the same answer, because cos² = 1- sin².

When a parameter, N, say, decays at a rate proportional to how much of it there is at a given time, we can write that  $$N=N_0e^{-\lambda t}$$  where  $$N_0$$  is the value of N at time t = 0 and  $$\lambda$$  is the decay rate.  The half-life,  $$\tau$$ , is related to  $$\lambda$$  by  $$\tau =\frac{ln2}{\lambda}$$ .

However, this might make more sense if you post a specific half-life question!

I assume this is meant to be  $$\frac{20xy^2-4xy}{4xy}$$  and you want to simplfy it.

The product 4xy is common to both terms in the numerator so you can write the expression as  $$\frac{4xy(5y-1)}{4xy}$$

The 4xy terms on the top and bottom cancel, so you are left with  $$5y-1$$

I suspect the questioner had positive integers in mind!

If your formula is correct then λ is

$${\frac{\left({\frac{{\mathtt{1.08}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sinh}}{\left({\frac{{\mathtt{1.08}}{\mathtt{\,\times\,}}{\mathtt{180}}}{{\mathtt{\pi}}}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cosh}}{\left({\frac{{\mathtt{1.08}}{\mathtt{\,\times\,}}{\mathtt{180}}}{{\mathtt{\pi}}}}\right)}\right)}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\frac{{\mathtt{1.08}}{\mathtt{\,\times\,}}{\mathtt{180}}}{{\mathtt{\pi}}}}\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{1.08}}{\mathtt{\,\times\,}}{\mathtt{180}}}{{\mathtt{\pi}}}}\right)}\right)}} = {\mathtt{3.872\: \!413\: \!402\: \!889\: \!445\: \!2}}$$

To get -0.56 you must have used 1.08*1 radians inside the trigonometric functions, but with those functions set to degree mode.  You need to (i) set the trig functions to radian mode, or (ii) convert radians to degrees in the trig functions (as I've done above).

You don't specify the orientation of the barge, nor along which diagonal the person is walking!  However, making the assumptions that the long direction is facing from one bank to the other, and that the person is walking with the stream, as illustrated below, we have the following vector addition problem:

The angle at which the person is walking relative to the barge is θ = tan^-1(48/20).

The x-direction velocity of the person relative to the stream is vx = 3.8 + 5*cos(θ) km/hr

The y-direction velocity of the person relative to the stream is vy = 4.5 + 5*sin(θ) km/hr

So you can use Pythagoras theorem to find the net velocity of the person relative to the stream (I'll leave that to you).

If the barge is oriented differently and/or the person is walking along the other diagonal you will have to make changes accordingly.

(With the situation as in my diagram I get the net velocity of the person relative to the stream as 10.763 km/hr) 

Speed is the rate of change of distance with time.  Differentiating L(t) = At - Bt² with respect to time (t) we get speed, S as

S(t) = dL(t)/dt = A - 2Bt

When t = 0, S(0) = 100km/hr, so:

100 = A - 2B*0 

or A = 100 km/hr.

when S(t) = 10 km/hr we have

10 = 100 - 2*90*τ   (where I've used τ  for the time at which speed is 10 km/hr).

Rearranging this we get

τ = (100 - 10)/(2*90) = 1/2

so τ = 1/2 an hour or 30 minutes.  (That's a long time to slow down!!).