Alan

Yes, I get 295.331 as well.

The original form of the denominator shows it is positive for all values of x (even when x is negative).  Therefore, if you take x outside the square root sign, you must still keep the denominator positive, so if x is negative you must have -x, instead of just x, to keep it positive.

To convince yourself that this is the right thing to do look at the following graph:

In the second figure below, how was 11.2v found?

The voltage drop from B1 to B2 is 28 - 7 = 21 volts

This is split into a drop across a 4 ohm resistor and a 1 ohm resistor in series with it.  This means the 21 volt  drop is divided in proportion to these two resistances.  

Let's call the voltage in between the two resistors V1.  Then (28 - V1)/(28 - 7) = R1/(R1 + R2)

(28 - V1)/21 = 4/(4 + 1) 

28 - V1 = 21*4/5

V1 = 28 - 21*4/5

$${\mathtt{V1}} = {\mathtt{28}}{\mathtt{\,-\,}}{\frac{{\mathtt{21}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{5}}}} \Rightarrow {\mathtt{V1}} = {\mathtt{11.2}}$$

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Why was 10 in the divide both sides by 10 step chosen? 

It's usually better to work with small numbers rather than large ones if possible, and it was clear that all the constants were multiples of 10 (they all ended in zero), so I divided by 10.  This step wasn't essential, the process works without doing this, but it just made the equations easier to view in my opinion.

I didn't understand this step. Please explain it- I looked at t for 5 minutes. Maybe I should look longer?

Multiply (3) by 4 and (4) by 5

4 = 20*I1 - 16*I2        ...(5)

-10 = -20*I1 + 30*I2  ...(6)

 The purpose of this step is to make the magnitudes of the coefficients of the I1 terms the same in both equations.  This means that the two equations could then be added together term by term and the I1 terms would disappear, leaving I2 as the only unknown. 

An alternative would have been to multiply (3) by 6 and (4) by 4.  In this case the magnitudes of the coefficients of the I2 term would have been the same, and I2 could have been eliminated, leaving I1 as the only unknown.

does this work for all silmultaneous equatoins, only those with 2 unknowns or only where one unknown will cancle? If it works for all i am very grateful. $_$

It will work for more than two equations, but only one term at a time will be eliminated.  Therefore the process must be repeated several times until you get down to a single unknown.

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Here's yet another way:

tanx = 2 implies sinx = 2cosx  ...(1)

cos2x = cos²x - sin²x.     Using (1) this is  cos2x = -3cos²x  ...(2)

sin²x + cos²x = 1.     Using (1) this is 5cos²x = 1   which means that  cos²x = 1/5   ...(3)

Put (3) in (2) to get   cos2x = -3/5

Hi Melody.  No, no-one knows how to generate an analytical solution to this indefinite integral as far as I'm aware.  (If it were known, I strongly suspect Mathematica (Wolfram Alpha) would be able to do it.)

It's sort of like trying to integrate a Gaussian (e^{-x^2}) only worse!  Actually, it was the Gaussian I had in mind with my first answer.  The difference here is that the integral from any finite limit to ∞ is likely to be infinite (in the case of the Gaussian it is finite).

This is essentially asking to integrate k^(x²), where k is a constant.  There is no analytical solution to this in general unless the limits are -∞ to ∞ (or one of these two limits is replaced by 0).  For other limits a numerical solution can be obtained (but this requires knowing what the limits are).

Stu, if you are unhappy about solving with matrices, you can always solve the two equations for I1 and I2 in the following way:

10 = 50*I1 - 40*I2      ...(1)

-20 = -40*I1 + 60*I2  ...(2)

Divide through both equations by 10

1 = 5*I1 - 4*I2         ...(3)

-2 = -4*I1 + 6*I2     ...(4)

Multiply (3) by 4 and (4) by 5

4 = 20*I1 - 16*I2        ...(5)

-10 = -20*I1 + 30*I2  ...(6)

Add (5) and (6) term by term

4 - 10 = 0 -16*I2 + 30*I2   Simplify

-6 =  14*I2

Divide both sides by 14

I2 = -6/14 = -3/7  

Put this back into (3)

1 = 5*I1 - 4*(-3/7) Simplify

1 = 5*I1 + 12/7

Subtract 12/7 from both sides

1- 12/7 = 5*I1  Simplify

-5/7 = 5*I1

Divide both sides by 5

I1 = -1/7

Or in terms of decimals

$${\mathtt{I1}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{7}}}} \Rightarrow {\mathtt{I1}} = -{\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$

$${\mathtt{I2}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{7}}}} \Rightarrow {\mathtt{I2}} = -{\mathtt{0.428\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

However, I think this is a poorly expressed question, as the definition of the currents changes within the question.  It would have been better to call the clockwise looping currents I3 and I4, say, to avoid confusion with the currents marked on the lines (although I1 is the same in both cases, I2 is not).

You have a right-angled triangle in which the height (the item you want to find) is opposite the known angle.  You also know the adjacent length to the angle.

This means you should use the tan function;  tan(angle) = opposite/adjacent.

Here:  tan(62°) = height/15

Multiply both sides by 15:   height = 15*tan(62°) ft

$${\mathtt{height}} = {\mathtt{15}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{62}}^\circ\right)} \Rightarrow {\mathtt{height}} = {\mathtt{28.210\: \!896\: \!980\: \!19}}$$

So height = 28 ft to the nearest foot.

Because sin can only take values between ±1 the values of tan(ln(x)) must lie in this range for the function to be defined.  This makes the function defined only between certain values of x.  See the graphs below:

There are an infinite number of such regions (e^7pi/4 < x < e^9pi/4, for example).

Which is the region of interest?