# Tutor profile: Natalie J.

## Questions

### Subject: Pre-Calculus

Find the domain and range for $$ f(x) = \sqrt{9 - x^{2}} $$.

Domain: f is defined for $$ \sqrt{9 - x^{2}} \geq$$ 0 or $$(3-x)(3+x) \geq 0$$. This separates out to $$3-x \geq 0$$, so $$x\leq3$$, and $$3+x \geq 0$$, so $$x\geq-3$$. So the domain is $$-3\leq x \leq 3$$. Range: Since $$y = \sqrt{9-x^{2}}$$ gives $$y^{2} = 9 - x^{2}$$ or $$y^{2} + x^{2} = 9$$. f(x) is the top half of the circle $$y^{2} + x^{2} = 9$$ so the range of f is the set of all y-coordinates of points on the semicircle, which is [0,3].

### Subject: Microeconomics

Consider the market supply of cereal. Indicate whether the following will cause a movement along the supply curve for cereal or a shift of the supply curve: a. An increase in the number of producers. b. A decrease in the price of oats (used in the production of cereal). c. An increase in the price of cereal.

a. Answer: this will cause a shift in the supply curve. An increase in number of producers means an increase in the market supply. b. Answer: this will cause a shift as well. A decrease in the price of oats means production is cheaper, so producers can make more of the good for the same cost, thus increasing supply. c. Answer: an increase in the price of cereal is a movement along the line, as it only causes a change in the quantity supplied. Remember that the supply curve is graphed with price on the y-axis and quantity of goods on the x-axis. A change in either will only cause a movement along the line.

### Subject: Algebra

Solve the equation: -0.25 x + 1.3 = -0.55 x - 0.2

The idea is to get x by itself on one side of the equation. Start by adding 0.55x to both sides: -0.25x + 1.3 + 0.55x = -0.55x - 0.2 + 0.55x Becomes: 0.30x + 1.3 = - 0.2 Then subtract 1.3 from both sides: 0.30x + 1.3 - 1.3 = -0.2 - 1.3 Becomes: 0.30x = - 1.5 Finally, divide both sides by 0.30: 0.30x/0.30 = - 1.5/0.30 Becomes: x = -5

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