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1. Find the distance from $(1,2)$ to $(-1,3)$.
Solution: Using the distance formula, we get $$d((1,2),(-1,3)) = \sqrt{(1 - (-1))^2 + (2-3)^2} = \sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt{5}.$$

2. Find the midpoint of the segment between $(-1,1)$ and $(5,3)$.
Solution: The midpoint is $$\left( \dfrac{-1+5}{2}, \dfrac{1+3}{2} \right) = \left( \dfrac{4}{2}, \dfrac{4}{2} \right) = (2,2).$$

3. Find all $x$ and $y$ intercepts of the equation $3y=6-4x$.
Solution: To find $x$-intercept, set $y=0$ and solve the resulting equation: $0=6-4x$. This yields solution $x=\dfrac{3}{2}$ and hence we have $x$-intercept $\left( \dfrac{3}{2},0 \right)$ (remember, intercepts are points!) To find the $y$-intercept, set $x=0$ and solve the resulting equation: $3y=6$ to get $y=2$. Hence the $y$-intercept is $\left( 0 ,2 \right)$.

4. Let $f(x)=2x+3$ and $g(x)=x^2+1$. Find $f(g(x))$ and $g(f(x))$.
Solution: First compute $f(g(x))$: $$f(g(x)) = f(x^2+1) = 2(x^2+1)+3 = 2x^2+2+3 = 2x^2+5.$$ Now compute $g(f(x))$: $$g(2x+3) = (2x+3)^2 + 1 = 4x^2 + 12x + 9 + 1 = 4x^2+12x+10.$$

5. Pictured below is an interval:

Express this interval using
a.) interval notation, and
b.) an inequality.
Solution: In interval notation, we write $[1,3)$. As an inequality we write $1 \leq x < 3$.