All the equations can be graphed and solved by graphing. They can also be solved using various other methods
R for ?
14) 96x1=24r
96=24r
r= 4
Answer:
I pretty sure the answer is Nervous
Answer:
b i think
Step-by-step explanation:
Answer:


Step-by-step explanation:
In single-variable calculus, the difference quotient is the expression
,
which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).
This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.
.
Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as <em>h</em> approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.
Therefore,

