5.65 in decimal and 113/20

To solve for n, we have to isolate n. To do so, we move all the terms that are not n to one side of the equation, and leave n on the other side.

Equation: n + 5/16 = -1
Subtract 5/16 on both sides to bring it to the right side of the equation.


Answer:
The answer is below
Step-by-step explanation:
The diameter of a tire is 2.5 ft. a. Find the circumference of the tire. b. About how many times will the tire have to rotate to travel 1 mile?
Solution:
a) The circumference of a circle is the perimeter of the circle. The circumference of the circle is the distance around a circle, that is the arc length of the circle. The circumference of a circle is given by:
Circumference = 2π × radius; but diameter = 2 × radius. Hence:
Circumference = π * diameter.
Given that diameter of the tire = 2.5 ft:
Circumference of the tire = π * diameter = 2.5 * π = 7.85 ft
b) since the circumference of the tire is 7.85 ft, it means that 1 revolution of the tire covers a distance of 7.85 ft.
1 mile = 5280 ft
The number of rotation required to cover 1 mile (5280 ft) is:
number of rotation = 
X
4
−34x
2
+225=0
2 Factor
x
4
−
34
x
2
+
225
x
4
−34x
2
+225.
(
x
2
−
25
)
(
x
2
−
9
)
=
0
(x
2
−25)(x
2
−9)=0
3 Solve for
x
x.
x
=
±
5
,
±
3
x=±5,±3
A(-7,-4) B(-2,0)
√[(x'-x)^2+(y'-y)^2]
√(-2-(-7)^2+(0-(-4)^2
√(5^2)+(4^2)
√25+16
√41
the distance is approximately 6.4 units