F(x)=\begin{cases}x^2-1 & \text{ for }\hspace{10px} x<0 \\ -4 & \text{ for }\hspace{10px} x=0 \\ \frac{3}{2}x+1 & \text{ for }\hspace{10px} 0< x \leq 4\end{cases} f(x)= ⎩ ⎪ ⎨ ⎪ ⎧ x 2 −1 −4 2 3 x+1 for x<0 for x=0 for 0

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.

M.xyz =radians. covert this radian measure to its equivalent measure in degrees.

If abcde is reflected over the x-axis and then translated 3 units left, what are the new coordinates d?