Answer:
The standard form of the equation for this line can be:
l: 2x + 5y = -15.
Step-by-step explanation:
Start by finding the slope of this line.
For a line that goes through the two points 
and 
,
.
For this line,
.
Find the slope-point form of this line's equation using
, and- The point
(using the point 
should also work.)
The slope-point form of the equation of a line
- with slope
and - point
should be 
.
For this line,
The equation in slope-point form will be
.
The standard form of the equation of a line in a cartesian plane is
where
, 
, and 
are integers. 
.
Multiply both sides of the slope-point form equation of this line by 
:
.
Add 
to both sides of the equation:
.
Therefore, the equation of this line in standard form is .
Answer:
Slope Intercept is the equation of a straight line in the form y = mx + b where m is the slope of the line and b is its y-intercept.
Step-by-step explanation:
See the attached picture with letters to better understand the problem
we know that
the figure is a regular hexagon
A regular hexagon has:
<span>Interior Angles of <span>120°
</span></span>so
in the triangle ABC
BC=2 units
∠ABC=30°
∠BAC=60°
tan 30°=AC/BC
solve for AC
AC=BC*tan 30°-------> AC=2*(√3)/3
<span>Apothem is equal to AC
</span>Apothem=(2/3)√3
the answer is(2/3)√3
Multiply 6 by 9 (whole number by denominator)
then add that to 15 (the product to the numerator.
Answer:
4x+2h
Step-by-step explanation:
The average rate of change of a continuous function,
f
(
x
)
, on a closed interval
[
a
,
b
]
is given by
f
(
b
)
−
f
(
a
)
b
−
a
So the average rate of change of the function
f
(
x
)
=
2
x
2
+
1
on
[
x
,
x
+
h
]
is:
A
r
o
c
=
f
(
x
+
h
)
−
f
(
x
)
(
x
+
h
)
−
(
x
)
=
f
(
x
+
h
)
−
f
(
x
)
h
...
.
.
[
1
]
=
2
(
x
+
h
)
2
+
1
−
(
2
x
2
+
1
)
h
=
2
(
x
2
+
2
x
h
+
h
2
)
+
1
−
2
x
2
−
1
h
=
2
x
2
+
4
x
h
+
2
h
2
−
2
x
2
h
=
4
x
h
+
2
h
2
h
=
4
x
+
2
h
Which is the required answer.
Additional Notes:
Note that this question is steered towards deriving the derivative
f
'
(
x
)
from first principles, as the definition of the derivative is:
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
This is the function we had in [1], so as we take the limit as
h
→
0
we get the derivative
f
'
(
x
)
for any
x
, This:
f
'
(
x
)
=
lim
h
→
0
4
x
+
2
h
=
4
x
, and
, and
.
