It’s D bc Dt= 15x3 and it’s not that hard to solve
Answer:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's find the answer.
Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.
Equations:
Eq. 1 : x1 + 2x2 + 5x3 = 5
Eq. 2 : x1 + x2 + x3 = 6
E1. 3 : 4x1 + 6x2 + 5x3 = 7
Coefficients for x1 ; x2 ; x3
From eq. 1 : 1 ; 2 ; 5
From eq. 2 : 1 ; 1 ; 1
From eq. 3 : 4 ; 6 ; 5
So matrix A is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D)
And the vector of vriables (X) is:
![\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D)
Now we can find the resulting vector (B) using the 'resulting values' from each equation:
![\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
In conclusion, AX=B is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Answer:
Step-by-step explanation:
A delivery of 4 boxes will contain 2 more defective lightbulbs than a delivery of 2 boxes.
Answer:
5 + 25 + 75 = 105 dollars
Step-by-step explanation:
The easiest way is to just take this day by day.
day 1: 5 dollars
day 2: day1^2 = 5^2 = 25
day 3: 3 * day2 = 3 * 25 = 75
Then sum them: 5 + 25 + 75
<span>B.4/7
You can check by taking 21 * 4/7 = 12 which is figure 2
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