The complete question is
Find the volume of each sphere for the given radius. <span>Round to the nearest tenth
we know that
[volume of a sphere]=(4/3)*pi*r</span>³
case 1) r=40 mm
[volume of a sphere]=(4/3)*pi*40³------> 267946.66 mm³-----> 267946.7 mm³
case 2) r=22 in
[volume of a sphere]=(4/3)*pi*22³------> 44579.63 in³----> 44579.6 in³
case 3) r=7 cm
[volume of a sphere]=(4/3)*pi*7³------> 1436.03 cm³----> 1436 cm³
case 4) r=34 mm
[volume of a sphere]=(4/3)*pi*34³------> 164552.74 mm³----> 164552.7 mm³
case 5) r=48 mm
[volume of a sphere]=(4/3)*pi*48³------> 463011.83 mm³----> 463011.8 mm³
case 6) r=9 in
[volume of a sphere]=(4/3)*pi*9³------> 3052.08 in³----> 3052 in³
case 7) r=6.7 ft
[volume of a sphere]=(4/3)*pi*6.7³------> 1259.19 ft³-----> 1259.2 ft³
case 8) r=12 mm
[volume of a sphere]=(4/3)*pi*12³------>7234.56 mm³-----> 7234.6 mm³
Answer:
10
Step-by-step explanation:
d = √(x2 - x1)² + (y2 - y1)²
d = √(-8 - -2)² + (9 - 1)²
d = √(-6)² + (8)²
d = √(36) + (64)
d = √100
d = 10
Answer:
12xy(3y-4x)
Step-by-step explanation:
Answer:
![x^2+y^2-5y+x=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-5y%2Bx%3D0)
Step-by-step explanation:
Let
be the equation of the circle. We know that
must be iqual
, due to the canonical equation of the circle is
, but we can think
as an unknown for the moment.
In order to determine the equation of the circle we have to find the coefficients
. In addition, we have that
are point on this circle. Furthermore, If
is any point in circle, then the following equations are satisfied:
![\begin{cases}a(x^2+y^2)+bx+cy+d=0\\a(2^{2}+3^{2})+b(2)+c(3)+d=0\\a((-3)^{2}+2^{2})+b(-3)+c(2)+d=0\\a(0^{2}+0^{0})+b(0)+c(0)+d=0\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da%28x%5E2%2By%5E2%29%2Bbx%2Bcy%2Bd%3D0%5C%5Ca%282%5E%7B2%7D%2B3%5E%7B2%7D%29%2Bb%282%29%2Bc%283%29%2Bd%3D0%5C%5Ca%28%28-3%29%5E%7B2%7D%2B2%5E%7B2%7D%29%2Bb%28-3%29%2Bc%282%29%2Bd%3D0%5C%5Ca%280%5E%7B2%7D%2B0%5E%7B0%7D%29%2Bb%280%29%2Bc%280%29%2Bd%3D0%5Cend%7Bcases%7D)
Here the unknowns are
. According with a result of linear algebra, the system has a nontrivial solution if and only if the determinant of the matrix of coefficients is zero, then we have that
![\left \lvert \begin{array}{cccc}x^{2}+y^{2}&x &y&1\\(2^{2}+3^{2})&2&3&1\\((-3)^{2}+2^{2})&-3&2&1\\(0^{2}+0^{2})&0&0&1 \end{array} \right \rvert =0](https://tex.z-dn.net/?f=%5Cleft%20%5Clvert%20%5Cbegin%7Barray%7D%7Bcccc%7Dx%5E%7B2%7D%2By%5E%7B2%7D%26x%20%26y%261%5C%5C%282%5E%7B2%7D%2B3%5E%7B2%7D%29%262%263%261%5C%5C%28%28-3%29%5E%7B2%7D%2B2%5E%7B2%7D%29%26-3%262%261%5C%5C%280%5E%7B2%7D%2B0%5E%7B2%7D%29%260%260%261%20%5Cend%7Barray%7D%20%5Cright%20%5Crvert%20%3D0)
Calculating this determinant we obtain the equation:
![13x^{2}+13y^2-65y+13x=0](https://tex.z-dn.net/?f=13x%5E%7B2%7D%2B13y%5E2-65y%2B13x%3D0)
Simplifying:
![x^2+y^2-5y+x=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-5y%2Bx%3D0)