Answer:
ΔV = 0.36π in³
Step-by-step explanation:
Given that:
The radius of a sphere = 3.0
If the measurement is correct within 0.01 inches
i.e the change in the radius Δr = 0.01
The objective is to use differentials to estimate the error in the volume of sphere.
We all know that the volume of a sphere
![V = \dfrac{4}{3} \pi r^3](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cdfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E3)
The differential of V with respect to r is:
![\dfrac{dV}{dr }= 4 \pi r^2](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdr%20%7D%3D%204%20%5Cpi%20r%5E2)
dV = 4 πr² dr
which can be re-written as:
ΔV = 4 πr² Δr
ΔV = 4 × π × (3)² × 0.01
ΔV = 0.36π in³
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 = ![\frac{5280\ ft}{7.85\ ft\ per\ rotation}=672\ rotations\\\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B5280%5C%20ft%7D%7B7.85%5C%20ft%5C%20per%5C%20rotation%7D%3D672%5C%20rotations%5C%5C%5C%5C%5C%5C)
I believe it is 3x+21 but i could be wrong :/ my explaination is -3 • -X would be +3x because it was distributed and -3 • -7 would be +21
Answer:The answer is 183, hope I helped.
Answer:
13,000
Step-by-step explanation: 10% of 31,000= 3,100 x 6= 18,600
31,000-18,600=13,000