Respuesta :
The original volume of the ballon, with radius r, is given by
V₀ = (4/3)πr³ in²
If the radius increases to (r+5) because of air input, the new volume is
V₁ = (4/3)π (r+5)³ in²
The difference in volume is the amount of air used to inflate the balloon.
The volume of air is
V = V₁ - V₀
= (4/3)π [r³ + 3r²(5) + 3r(5²) + 5³ - r³]
= (4/3)π (15r² + 75r + 125) in³
Answer: (4/3)π (15r² + 75r + 125) in³
V₀ = (4/3)πr³ in²
If the radius increases to (r+5) because of air input, the new volume is
V₁ = (4/3)π (r+5)³ in²
The difference in volume is the amount of air used to inflate the balloon.
The volume of air is
V = V₁ - V₀
= (4/3)π [r³ + 3r²(5) + 3r(5²) + 5³ - r³]
= (4/3)π (15r² + 75r + 125) in³
Answer: (4/3)π (15r² + 75r + 125) in³
Expression that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 5 inches will be [tex]\rm V = \dfrac{4}{3}\times \pi \times (15r^2+75r+125)\;inches^3[/tex] and this can be determine by taking the difference between the final volume and original volume.
Given :
A spherical balloon with radius r inches has volume, [tex]\rm V(r) = \dfrac{4}{3}\pi r^3[/tex].
The volume of the balloon when radius is r, that is, the original volume:
[tex]\rm V_1 = \dfrac{4}{3}\pi r^3[/tex]
The volume of the balloon when radius is (r + 5), that is:
[tex]\rm V_2 = \dfrac{4}{3}\times \pi\times (r+5)^3[/tex]
[tex]\rm V_2 = \dfrac{4}{3}\times \pi\times (r^3+5^3+(3\times r \times 5 (r+5)))[/tex]
[tex]\rm V_2 = \dfrac{4}{3}\times \pi\times (r^3+125+15r (r+5))[/tex]
[tex]\rm V_2 = \dfrac{4}{3}\times \pi\times (r^3+125+15r^2 +75r)[/tex]
An expression that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of (r + 5) inches will be:
[tex]\rm V = V_2 -V_1[/tex]
[tex]\rm V = \dfrac{4}{3}\times \pi \times (r^3+15r^2+75r+125-r^3)[/tex]
[tex]\rm V = \dfrac{4}{3}\times \pi \times (15r^2+75r+125)\;inches^3[/tex]
For more information, refer the link given below:
https://brainly.com/question/1578538
