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
The differential of V with respect to r is:
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
5x 3y^3 2z
I know it is in standard form because there are no more like terms.
Part B: Polynomials are always closed under multiplication. Unlike with addition and subtraction, both the coefficients and exponents can change. The variables and coefficients will automatically fit in a polynomial. When there are exponents in a multiplication problem, they are added, so they will also fit in a polynomial.
Or,
Refer to the photos:
hope this help :)
