Alan

On the scientific calculator on this site the antilog function is the key immediately to the right of the log function.  The default keys are log₁₀ and 10^x, but if you hover the cursor over either of these you also get ln and e^x, and log₂ and 2^x.

 NOW can someone tell me how to write limit as h tends to 0 in latex please?

The following:  

$$\lim h\rightarrow0\frac{3x+h}{x+h}$$ (in Texmaker go to Math/Math functions to get \lim)gives:

$$$$\lim h\rightarrow0\frac{3x+h}{x+h}$$$$

but I don't know how to get the h->0 to go just beneath lim. We need a LaTeX expert!

The left-hand side is an upward curving quadratic and the right-hand side is a straight line with a positive slope that cuts through the quadratic at two points.  To see this draw curves of y = x²+x-12 and y = 4x on the same graph

Treat it as an equality first and find the two values of x that solve the equality, then the values of x in between those two give the range in which the inequality holds.

There is a Chi-square test function in Excel:  

CHISQ.TEST(actual_range,expected_range)

Search CHISQ in Excel's help file.

Energy of photons is proportional to frequency, so inversely proportional to wavelength.  Therefore the ratio of ultraviolet to average energy is (1/3*10^-7)/(1/5.6*10^-7) = 5.6/3 (because the 10^-7s cancel).  So ultraviolet energy is 

$${\frac{{\mathtt{5.6}}}{{\mathtt{3}}}} = {\mathtt{1.866\: \!666\: \!666\: \!666\: \!666\: \!7}}$$times greater than the average.

Let the common length be x. Then the perimeter is 2x + (2x-2) or 4x-2, so we must have 4x-2 = 1.8

$${\mathtt{x}} = {\frac{\left({\mathtt{1.8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}{{\mathtt{4}}}} = {\mathtt{x}} = {\mathtt{0.95}}$$metres

 The third side is then 2*0.95 - 2 which is less than zero!  So there is a problem with the question specification!

With log to the base 10 this is saying log₁₀x = 0.3142, what is x?

From definition of logarithm this means x = 10^0.3142 so:

$${\mathtt{x}} = {{\mathtt{10}}}^{{\mathtt{0.314\: \!2}}} = {\mathtt{x}} = {\mathtt{2.061\: \!579\: \!086\: \!952\: \!637\: \!4}}$$

This is 2.0616 to four decimal places.

This requires conservation of energy.  Kinetic energy at the bottom of the hill equals the sum of kinetic and potential energies at the top.

KE at bottom = (1/2)mv₁², where m is mass and v₁ is velocity at bottom (35.9ms^-1)

PE at top = mgh, where g is gravitational acceleration (9.81ms^-2) and h is height if hill (17.9m)

KE at top = (1/2)mv₂²,  so

(1/2)mv₂² + mgh = (1/2)mv₁² 

The mass cancels throughout, so

(1/2)v₂² +9.81*17.9 = (1/2)35.9²,

Multiply throughout by 2

v₂² + 2*9.81*17.9 = 35.9²,  

See if you can take it from here!

I take this to mean a² - 4 = 5, inwhich case a² = 9, so there are two possibilities for a, namely a = 3 and a = -3. 

This isn't a quadratic equation, it's a linear one. You need a z² term somewhere to make it quadratic.