Here is the order
- 6 2/3 ,- 1.3 , 29/8 , 7
Answer: ∆V for r = 10.1 to 10ft
∆V = 40πft^3 = 125.7ft^3
Approximate the change in the volume of a sphere When r changes from 10 ft to 10.1 ft, ΔV=_________
[v(r)=4/3Ï€r^3].
Step-by-step explanation:
Volume of a sphere is given by;
V = 4/3πr^3
Where r is the radius.
Change in Volume with respect to change in radius of a sphere is given by;
dV/dr = 4πr^2
V'(r) = 4πr^2
V'(10) = 400π
V'(10.1) - V'(10) ~= 0.1(400π) = 40π
Therefore change in Volume from r = 10 to 10.1 is
= 40πft^3
Of by direct substitution
∆V = 4/3π(R^3 - r^3)
Where R = 10.1ft and r = 10ft
∆V = 4/3π(10.1^3 - 10^3)
∆V = 40.4π ~= 40πft^3
And for R = 30ft to r = 10.1ft
∆V = 4/3π(30^3 - 10.1^3)
∆V = 34626.3πft^3
Answer:
look at this
o(*≧□≦)o
Step-by-step explanation:
MATH-WAY
Answer:
The dilation is an enlargement by 3
Step-by-step explanation:
I took geometry lasy yr
Answer:
t = 13.2 s
Step-by-step explanation:
Given that,
An athlete ran the 100 meter sprint in 13.245 seconds. We need to round off the time to the nearest tenth.
Here, t = 13.245 s
In order to round off to the nearest tenth, thousandths place and hundredth place gets eliminated.
So, t = 13.2 s
So, time is 13.2 seconds.