On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3) crosses the x-axis at (negative 4, 0) and crosses the y-axis at (0, negative 3). Which statement is true about the graphed function? F(x) 0 over the interval (–∞, –3) F(x) > 0 over the interval (–∞, –4)

f(x) > 0 over the interval

Step-by-step explanation:

If  f(x)  is a continuous function,  and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.

Then the function was above the x axis (larger than zero) from , until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.

There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from to -3, since between the  values -4 and -3 the function adopts negative values.

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.

i need a picure

step-by-step explanation: