So to fine slope you would use the formula down below:

rise/run

So use a graphed point, 0, -5 and you rise or count up quadrants up to a point and then horizontally move to when you find that point.

So from 0,-5 go up 9 vertically, and you would be on the 4

Go horizontal 3 spots and your on a designated point.

So the rise is four and the run is 3

So 4/3 is the slope

In the y= Mx + b equation you would set the equation like this:

y= 4/3 + -5

The m in this formula stands for the slop and the b stands for the y-intercept

The y-intercept is the point that is on the y-axis and where it starts.

6 0

1 year ago

Let A(t) be the area of a circle with radius r(t), at time t in min. Suppose the radius is changing at the rate of drdt=6 ft/min

mylen [45]

Answer:

The rate of change is $108\pi$ ft^(2)/min

Step-by-step explanation:

The area of a circle is given by the following equation:

$A(t) = \pi r^{2}$

To solve this question, we have to realize the implicit differentiation in function of t. We have two variables, A and r. So

$\frac{dA(t)}{dt} = 2\pi r \frac{dr}{dt}$

We have that:

$\frac{dr}{dt} = 6, r = 9$ .

We want to find $\frac{dA}{dt}$

$\frac{dA(t)}{dt} = 2\pi*9*6$

$\frac{dA}{dt} = 108\pi$

Since the area is in square feet, the rate of change is in ft^(2)/min.

So the rate of change is $108\pi$ ft^(2)/min

5 0

3 years ago

Going through York Canal, he drives 5 miles in 15 minutes. Later on as he crosses Stover Lake, he drives 30 minutes at the same

SVETLANKA909090 [29]

When going through York Canal, his speed is 1 mile per 3 minutes.

If going by this speed while crossing Stover Lake, he would drive 10 miles in 30 minutes.

Therefore, he drove 10 miles.

5 0

3 years ago