We know that sine functions of the form f(x)=a*sin(kx) are odd functions because f(-x)=a*sin(-kx)=-a*sin(kx)=-f(x) We also know that cosine functions of the form g(x)=b*cos(kx) are even functions because g(-x)=b*cos(-kx)=b*cos(kx)=g(x)
Out of the four choices, (a)F(x)=2sin(x/2+pi/2)=2cos(x/2) => even function (b)F(x)=2sin(x/2)=a*sin(kx) where a=2, k=1/2 => odd function (c)F(x)=2cos(2x) => even function (d)F(x)=cos(x+pi)=-cos(x) = b*cos(kx) where b=-1, k=1, => even function
The sine function is always odd. That means you can elimiinate both cosine functions here. Another requirement for an odd function is that the graph goes thru the origin (0,0). This is true of Graph B.
In the question, it says that her coach wants her to run 2 1/2 TIMES the original 3/4 she had ran. Therefore, you would want to make the fractions into improper fractions, and then multiply, which is how you would get 13/4 * 5/2.
