Answer:
The value of x is: x > -2
Step-by-step explanation:
The interval notation if you need it is (-2, ♾)
Answer:
156°
Step-by-step explanation:
The sum of the interior angles of a polygon is
sum = 180° (n - 2) ← n is the number of sides
Here n = 15 , then
sum = 180° × 13 = 2340°
interior angle =
=
= 156°
Answer:

Step-by-step explanation:
Let r represent Linda's walking rate.
We have been given that Linda can ride 9 mph faster than she can walk, so Linda's bike riding rate would be
miles per hour.

We have been given that Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles.


Since both times are equal, so we will get:

Therefore, the equation
can be used to solve the rates for given problem.
Cross multiply:





Therefore, Linda's walking at a rate of 3 miles per hour.
Linda's bike riding rate would be
miles per hour.
Therefore, Linda's riding the bike at a rate of 12 miles per hour.
Answers:
Answer for row one: 1
Answer for row two: 11
Answer for row three: 16
Answer for row four: 36
=========================================
Work Shown:
Whatever the x value is, we multiply by 5 and subtract off 14 to get the corresponding y value. This is following the order of operations PEMDAS
------------
If x = 3, then
y = 5*x - 14
y = 5*3 - 14 ..... note how x is replaced with 3
y = 15 - 14
y = 1
This means that when x = 3, the y value is y = 1.
So 1 goes in the box in the first row.
---------------
Repeat for x = 5
y = 5*x - 14
y = 5*5 - 14
y = 25 - 14
y = 11
We have 11 as the second answer.
--------------
Repeat for x = 6
y = 5*x - 14
y = 5*6 - 14
y = 30 - 14
y = 16
The third answer is 16.
--------------
Repeat for x = 10
y = 5*x - 14
y = 5*10 - 14
y = 50 - 14
y = 36
The last answer is 36.
f
'
(
x
)
=
1
(
x
+
1
)
2
Explanation:
differentiating from first principles
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
f
'
(
x
)
=
lim
h
→
0
x
+
h
x
+
h
+
1
−
x
x
+
1
h
the aim now is to eliminate h from the denominator
f
'
(
x
)
=
lim
h
=0
(
x
+
h
)
(
x
+
1
)−
x
(
x
+
h
+
1)
h
(
x
+
1
)
(
x
+
h
+
1
)
f
'
(
x
)
=
lim
h
→
0
x
2
+
h
x
+
x
+
h
−
x
2
−
h
x
−
x
h
(
x
+
1
)
(
x+h
+
1
)
f
'
(
x
)
=
lim
h
→
0
h
1
h
1
(
x
+
1
)
(
x
+
h
+1
)
f
'
(
x
)
=
1
(
x
+
1
)
2