Alan

The images don't show here (I guess you need to have an account at Physics Central before you can see them), so you'll need to specify the detail in words or find a way of inserting the images using the Tiny Pic method.

The centripetal force required to keep the car from skidding is given by

$$F=m\frac{v^2}{r}$$

where m is the mass of the car, v is it's speed and r is the curve radius.

The frictional force is

$$F=\mu mg$$

where μ is the coefficient of friction and g is gravitational acceleration.

So equating these two forces we have:

$$\frac{v^2}{r}=\mu g\\\\
v=\sqrt{\mu gr}$$

I'll let you put in the numbers.

My experience is that the scientific community in the UK adopted the US approach to billions and trillions a long time ago!

Melody said:  I'd never remember the alt number in a million years!

You would if you used it a lot!!

It is also possible that the equation in question is meant to be:

$$e^{-0.2x}-e^{-0.75x}=0.25$$ 

In this case there are two solutions as can be seen from the graph below:

These solutions can be obtained numerically.  A simple iterative approach is to write the equation in the following form:

$$x_{n+1}=\frac{-1}{0.2}ln(0.25+e^{-0.75x_n})$$ 

Then let x0 = 1, say, and use the formula above to iterate several times.  The result is:

x ≈ 6.812073

By rewriting the iteration formula as:

$$x_{n+1}=\frac{-1}{0.75}ln(e^{-0.2x_n}-0.25)$$

we can iterate to the other solution:

x ≈ 0.6023554654

From here you can get the degree symbol by selecting Insert, Special character and choosing the degree symbol ° (2nd row, fifth character from the right).

More generally, you can type Alt 0176 from the keyboard's numeric pad ° (you have to hold down the Alt key while typing the 0176).

Incidentally, Melody, to get out of superscript mode just select superscript again!

Functions that make use of angles (like sin and cos, for example) have to know if the angles they are working with are in degrees or radians.  If the calculator is in degree mode the functions automatically assume that numbers they are given are in degrees; if it is in radian mode the functions automatically assume that numbers they are given are in radians.

eg degree mode:   cos(5) = 0.9962   the 5 is assumed to be degrees

radian mode:  cos(5) = 0.2837  the 5 is assumed to be radians

You can't!  There are an infinite number of different sized right triangles with a given base.

I assume that by "row" you mean "in series" and by "beside" you mean "in parallel".  If that is the case then the following diagram illustrates your circuit:

Resistances in series simply add:  R24 = R2+R4 and R35 = R3 + R5

For resistances in parallel: 1/R2345 = (1/R24) + (1/R35)

Then R1 and R2345 are in series so Rtotal = R1 + R2345

The total current is then given by I = U/Rtotal

Can you take it from there?

(You should find that I ≈ 0.0224 A)