Answer:

Step-by-step explanation:
Assuming that the first exponent in the formula for the curve should be 3, not 2...

The derivative is

The slope of the tangent line at the point (-1, 2) is the value of the derivative at x = -1.

The slope of the normal line is the opposite reciprocal of the slope of the tangent line.

Using the Point-Slope form of a linear equation, the normal line is

Answer:
70
Step-by-step explanation:
it depends on the grade you make, so if you were to make a 65.47, your grade would be a 70, so any grade less than 70 should be the answer, hope this helps!
I don't have the ranges of numbers to see which range so clearly it's impossible to answer
Its b
2854/32 = 1427/16 = 89 3/16
+1,-1 and your other would be 1,1