I disagree with the student because the slope is the rate of change. The slope would be 10 because the rocks are increasing by 10 every month.
Answer:
<em>201.06 sq. ft.</em>
Step-by-step explanation:
Area of a circle is found by using the equation:

Radius = 1/2 (diameter) = 1/2 (16) = 8
So plug in:

A = 64
A = 201.06193
Rounded A= 201. 06
The slope-intercept form is
y=mx+b
, where m is the slope and b is the y-intercept.
y=mx+b
Find the values of
m
and
b
using the form
y
=
m
x
+
b
.
\m
=
2
b
=
−
3
The slope of the line is the value of
m
, and the y-intercept is the value of
b
.
Slope:
2
y-intercept:
(
0
,
−
3
)
Any line can be graphed using two points. Select two
x
values, and plug them into the equation to find the corresponding
y
values.
Tap for fewer steps...
Find the x-intercept.
Tap for more steps...
x-intercept(s):
(
3
2
,
0
)
Find the y-intercept.
Tap for more steps...
y-intercept(s):
(
0
,
−
3
)
Create a table of the
x
and
y
values.
x
y
0
−
3
3
2
0
m
=
2
b
=
−
3
The slope of the line is the value of
m
, and the y-intercept is the value of
b
.
Slope:
2
y-intercept:
(
0
,
−
3
)
Any line can be graphed using two points. Select two
x
values, and plug them into the equation to find the corresponding
y
values.
Tap for fewer steps...
Find the x-intercept.
Tap for more steps...
x-intercept(s):
(
3
2
,
0
)
Find the y-intercept.
Tap for more steps...
y-intercept(s):
(
0
,
−
3
)
Create a table of the
x
and
y
values.
x
y
0
−
3
3
2
0
I'm so sorry it layed out like this my computer is being st00pid
Answer:
(f+g)(2) = 4
(f-g)(4) = 8
(f ÷g)(2) = 7
(f x g)(1) = 0
Step-by-step explanation:
We are given these following functions:


(f+g)(2)

At 

Then
(f+g)(2) = 4
(f-g)(4)

At x = 4

Then
(f-g)(4) = 8
(f ÷g)(2)

At x = 2

Then
(f ÷g)(2) = 7
(f x g)(1)

Then

So
(f x g)(1) = 0