<span>the question does not have the options, but this does not limit its resolution
</span>
we know that
volume of the cone=(1/3)*pi*r²*h
where
r is the radius of the cone
h is the height of the cone
r=12 ft
Volume=1506.96 ft³
h=?
clear the variable h
v=(1/3)*pi*r²*h-----> multiply by 3
3*v=pi*r²*h----> divided by (pi*r²)
h=3*v/(pi*r²)------> h=3*1506.96/(pi*12²)----> h=9.99 ft---> h=10 ft
the answer is
h=10 ft
Answer:
y = 1/5x + 8
Step-by-step explanation:
slope = (change in y)/(change in x)
Slope = (8 -10)/(0 -10)
Slope = -2/-10
Slope = 1/5
Use y = mx + b substitute m = 1/5 and (0, 8)
8 = 1/5(0) + b
8 = 0 + b
8 = b
The equation of the equation y = 1/5x +8
Answer:
672 ways
Step-by-step explanation:
The total number of ways the friends were supposed to sit without restrictions would have been = 6!
= 720 possible ways.
If the girs should sit on the first or last sit
The possible arrangement= 2!4!
= 2*24
= 48
But now the two girls should be arranged in such a way that the don't sit on either the first or the last chair.
So the possible way now is= way without restriction - way if the girls are to sit at either first or last sit
= 720-48
= 672
So we are given the system:
Written in matrix form we get:
![\left[\begin{array}{cc}2&4\\6&3\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}8\\-3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C6%263%5Cend%7Barray%7D%5Cright%5D%20%0A%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20)
We compute the solution like this:
![ \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{cc}2&4\\6&3\end{array}\right] ^{-1} \left[\begin{array}{c}8\\-3\end{array}\right] \\= \left[\begin{array}{cc}-3&4\\6&-2\end{array}\right] \left[\begin{array}{c}8\\-3\end{array}\right] \dfrac{1}{18}\\= \left[\begin{array}{c}2\\-3\end{array}\right]](https://tex.z-dn.net/?f=%20%0A%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%0A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C6%263%5Cend%7Barray%7D%5Cright%5D%20%5E%7B-1%7D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%264%5C%5C6%26-2%5Cend%7Barray%7D%5Cright%5D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%5Cdfrac%7B1%7D%7B18%7D%5C%5C%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
The solution is :
![\left[\begin{array}{c}2\\-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
