# Tutor profile: Emma T.

## Questions

### Subject: Basic Math

**Best for Grades 4-7** The faces on a fair number die are labelled 1, 2, 3, 4, 5 and 6. You roll the die 12 times. How many times should you expect to roll a 1?

Since the dice are fair all the sides have an equal probability of coming up. There are 6 sides on a dice so each side has a 1/6th chance of coming up on any one roll. This means if we roll the dice 6 times we would expect each side to come up once (because 1/6 x 6 = 1). Now if roll the 6 more times we would expect each side to come up one more time. So now we have rolled the dice 12 times and each number has come up twice. We can do this quicker by multiplying the chance of 1 coming up on one roll by the total number of rolls, so 1/6 x 12 = 2.

### Subject: Political Science

What are the primary functions of the Canadian constitution?

Firstly, the constitution defines who has power in each area of government. Legislative power is the power to make laws or policies, executive power is the power to “execute” or administer the laws and policies, and judicial power is the power to settle questions about specific violations of law and to choose a suitable punishment from among those permitted in the relevant legislation for those found guilty. In addition, the constitution provides a division of powers between national and regional governments, in the Canadian case, this is primarily defining what is federal jurisdiction and what is provincial jurisdiction. Further, the constitution delineates the limits of governmental power and provides for an orderly way to make changes to the constitution.

### Subject: Calculus

Determine an equation of the tangent to the graph of $$f(x)=\frac{1}{x}$$ at the point where $$x=2$$

When $$x=2, f(x)=\frac{1}{2}$$ so we are looking for the tangent at the point $$(2,\frac{1}{2})$$ First, we must find the derivative of $$f(x)$$ AKA $$f'(x)$$ because this represents the slope of the tangent line To find our derivative $$f'(x)$$ we are going to use the equation $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ Next, we substitute in $$f(x+h) = \frac{1}{x+h}$$ and $$f(x)=\frac{1}{x}$$ $$f'(x)=\lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$ Next, we simplify our fraction. We want to make the denominater not just h so that when we take the limit (set h to 0) we get a proper fraction. $$f'(x)=\lim_{h \to 0} \frac{\frac{x}{x(x+h)}-\frac{x+h}{x(x+h}}{h}$$ $$f'(x)=\lim_{h \to 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}$$ $$f'(x)=\lim_{h \to 0} \frac{\frac{x-x-h}{x(x+h)}}{h}$$ $$f'(x)=\lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}$$ $$f'(x)=\lim_{h \to 0} \frac{-h}{(h)(x)(x+h)}$$ $$f'(x)=\lim_{h \to 0} \frac{-1}{(x)(x+h)}$$ Now we can take the limit by setting $$h=0$$ $$f'(x)= \frac{-1}{(x)(x+0)}$$ $$f'(x)= \frac{-1}{x^2}$$ This equation is now the slop of $$f(x)$$ at any point x, since we are looking for the tangent line at $$x=2$$ we need to substitute in $$x=2$$. $$f'(2)=-\frac{1}{2^2}=-\frac{1}{4}$$ So $$-\frac{1}{4}$$ is the slope, m, of our tangent line. Therefore using the equation of the tangent line $$y-f(a)=m(x-a)$$ we can find the equation of our tangent line. $$a=2, f(a)=\frac{1}{2}, m=-\frac{1}{4}$$ $$y-\frac{1}{2}=-\frac{1}{4}(x-2)$$ Which in standard form is $$x+4y-4=0$$

## Contact tutor

needs and Emma will reply soon.