Let x = Initial Price
If we increase x by 5%, we are adding 0.05x
Therefore, the new price = x + 0.05x = 1.05x
If the ticket has increased by £2.30, £2.30 is 5% of the initial price, or 0.05x
0.05x = 2.30
x = 2.30/0.05
x = 46
Therefore, the price of the ticket before the increase was £46
You can also check this backwards by doing 46*0.05 = 2.30
Answer:
(a) k'(0) = f'(0)g(0) + f(0)g'(0)
(b) m'(5) = 
Step-by-step explanation:
(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e., ![\frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Bf%28x%29%5Ctimes%20g%28x%29%5D%7D%7B%5Cmathrm%7Bd%7D%20x%7D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20f%28x%29%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5Ctimes%20g%28x%29%20%2B%20f%28x%29%5Ctimes%5Cfrac%7B%5Cmathrm%7Bd%7D%20g%28x%29%7D%7B%5Cmathrm%7Bd%7D%20x%7D)
this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>
on substituting the value x=0, we will get the value of k'(0)
{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}
(b) m(x) is a function of two functions f(x) and g(x) [
]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows : 
this will give <em>
</em>
on substituting the value x=5, we will get the value of m'(5).
{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}
{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }
{ NOTE :
}
By what I can tell, that is telling you what the measurement of rain was per month. For example, August had 10. That means that in the month of August there was 10 inches of rain.