R^2+(ab)^2= (ao)^2
ab=6
ao=11.7
Plug in
r^2+6^2=11.7^2
simplify
r^2+36= 136.89
-36 both sides
r^2=100.89
square root both sides
r= 10.04 rounded 10
Answer: 8 pizzas.
Step-by-step explanation:
4/5 of 10 is 8. You can multiply 4/5 X 10 to get 8. Therefore, 8 pizzas were eaten.
Answer:
33.33%
Step-by-step explanation:
the syrup has a total of 15kg of ingredients, 5kg of it is sugar, so 15/5 which will end as 33.33%
Answer with explanation:
1. The given equations are
3x -5 y=2
-x+2 y= 0
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-5%5C%5C-1%262%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%2C%5C%5C%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D)
Adj.A=Transpose of cofactor of Matrix A
![Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2](https://tex.z-dn.net/?f=Adj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C5%263%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20%7CA%7C%3D6-5%5C%5C%5C%5C%7CA%7C%3D1%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D4%2C%20y%3D2)
2.
The given equations are
x+y-z=2
x+z=7
2 x +y+z=13
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26-1%5C%5C1%260%261%5C%5C2%261%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Crightarrow%20X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Crightarrow%20X%3D%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%5Ctimes%20B%5C%5C%5C%5Ca_%7B11%7D%3D-1%2Ca_%7B12%7D%3D1%2Ca_%7B13%7D%3D1%2Ca_%7B21%7D%3D-2%2Ca_%7B22%7D%3D3%2Ca_%7B23%7D%3D1%2Ca_%7B31%7D%3D1%2Ca_%7B32%7D%3D-2%2Ca_%7B33%7D%3D-1%5C%5C%5C%5C%7CA%7C%3D1%5Ctimes%280-1%29-1%5Ctimes%281-2%29-1%5Ctimes%281-0%29%5C%5C%5C%5C%3D-1%2B1-1%5C%5C%5C%5C%7CA%7C%3D-1%5C%5C%5C%5CAdj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-2%261%5C%5C1%263%26-2%5C%5C1%261%26-1%5Cend%7Barray%7D%5Cright%5D)
![\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4](https://tex.z-dn.net/?f=%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CX%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5Ctimes%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D3%2Cy%3D3%2Cz%3D4)
Answer: During the second hour it could travel either 4,5,6,7,8,9 hours
Step-by-step explanation:
