The volume of the sphere of radius r is:
V1 = (4/3) * (pi) * (r ^ 3)
Where,
r: sphere radius:
The volume of the sphere of radius 0.3r is:
V2 = (4/3) * (pi) * ((0.3r) ^ 3)
Rewriting:
V2 = (4/3) * (pi) * (0.027 (r) ^ 3)
V2 = 0.027 (4/3) * (pi) * (r ^ 3)
V2 = 0.027V1
The difference is:
V1-V2 = V1-0.027V1 = V1 (1-0.027)
V1-V2 = 0.973 * (4/3) * (pi) * (r ^ 3)
Answer:
the difference in volume between a sphere with radius and a sphere with radius 0.3r is:
V1-V2 = 0.973 * (4/3) * (pi) * (r ^ 3)
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
9m(n)^3 +14m(n)^2
Step-by-step explanation:
6m(n)^3 - m(n)^2 + 3m(n)^3 + 15m(n)^2
=> 6m(n)^3 + 3m(n)^3 + 15m(n)^2 - m(n)^2
=> 9m(n)^3 +14m(n)^2
