It will be Net A and the S. A. = 12*14 + 5*14 + 2(0.5*5*12) + 14*13 = 480
The correct option is the second one.
Guessing it is a rectangle the perimeter is 2*0.625+2.45*2= 6.15
and the area is 0.625*2.45= <span>1.53125..... i think</span>
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:
0.1 , 0.2 , 0.05
Step-by-step explanation:
*each column adds to 1*
20 x 1/2 = 10
1/2 x 20 x 0.1 = 1
top left box = 0.1
; 5 x 2 x 0.1 = 1
1/2 x 10 = 5
; 0.2 x 1/2 x 10 = 1
bottom middle box = 0.2
; 5 x 1 x 0.2 = 1
; 2 x 10 x 0.05 = 1
middle right box = 0.05
; 20 x 1 x 0.05
Answer:
x = 8 is the answer
Step-by-step explanation:
In order to find the solution of Equations using Cramer's rule
we shall find matrices
D= ![\left[\begin{array}{ccc}5&-3\\1&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-3%5C%5C1%26-2%5Cend%7Barray%7D%5Cright%5D)
=
![D_{1} = \left[\begin{array}{ccc}4&-3\\-16&-2\end{array}\right]](https://tex.z-dn.net/?f=D_%7B1%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26-3%5C%5C-16%26-2%5Cend%7Barray%7D%5Cright%5D)
l D l = Determinant formed by coefficient of variables
Here l D l = 5X(-2) - 1X(-3)= -10+3 = -7
l D1 l = Determinant formed by replacing first column In D by numbers on the right side of the equations
Here l D1 l = 4X(-2)- (-16)X(-3)= -8-48 = -56
here x is given by 
x = 
x = 8
