Use the given information as (x, y) points.
You have
(1, 34)
(2, 50)
(3, 66)
(4, 82)
The difference between renting for 1 day and 2 days is 50 - 34 = 16.
The difference between renting for 2 days and 3 days is 66 - 50 = 16
The difference between renting for 3 days and 4 days is 82 - 66 = 16
For each day you rent you pay $16.
Now look at renting for 1 day.
The rental for 1 day is $16, but for 1 day you pay $34.
$34 - $16 = $18
There is an extra $18 in the rental for every day. The $18 is fixed.
That means, to rent a truck, you must pay a fixed amount of $18 plus $16 per day.
total cost = daily cost + fixed cost
total cost = 16d + 16
c = 16d + 18
Answer: speed v= s/t = (460 / 5,75 )km/h
Step-by-step explanation: s= 460 km, t= 5h 45 min = 5,75 h
Hey!~
The answer is 25/100.
0/100 (0) and 50/100 (1/2) : The middle is 1/4.
Hope this helps and have a great day!
-Lindsey
Answer:
108.36 ft2
Step-by-step explanation:
Use the facts we already know and plug them into the equation.
a=12.6(8.6)
If we multiply both factors, we get 108.36.
So, the rectangle is 108.36 ft2
Answer:
Matrix multiplication is not conmutative
Step-by-step explanation:
The matrix multiplication can be performed if the number of columns of the first matrix is equal to the number of rows of the second matrix
Let A with dimension mxn and B with dimension nxp represent two matrix
The multiplication of A by B is a matrix C with dimension mxp, but the multiplication of B by A is can't be calculated because the number of columns of B is not the number of rows of A. Therefore, you can notice that is not conmutative in general.
But even if the multiplication of AB and BA is defined (For example if A and B are squared matrix of 2x2) the multiplication is not necessary conmutative.
The matrix multiplication result is a matrix which entries are given by dot product of the corresponding row of the first matrix and the corresponding column of the second matrix:
![A=\left[\begin{array}{ccc}a11&a12\\a21&a22\end{array}\right]\\B= \left[\begin{array}{ccc}b11&b12\\b21&b22\end{array}\right]\\AB = \left[\begin{array}{ccc}a11b11+a12b21&a11b12+a12b22\\a21b11+a22b21&a21b12+a22b22\end{array}\right]\\\\BA=\left[\begin{array}{ccc}b11a11+b12a21&b11a12+b12a22\\b21a11+b22ba21&b21a12+b22a22\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da11%26a12%5C%5Ca21%26a22%5Cend%7Barray%7D%5Cright%5D%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db11%26b12%5C%5Cb21%26b22%5Cend%7Barray%7D%5Cright%5D%5C%5CAB%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da11b11%2Ba12b21%26a11b12%2Ba12b22%5C%5Ca21b11%2Ba22b21%26a21b12%2Ba22b22%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CBA%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db11a11%2Bb12a21%26b11a12%2Bb12a22%5C%5Cb21a11%2Bb22ba21%26b21a12%2Bb22a22%5Cend%7Barray%7D%5Cright%5D)
Notice that in general, the result is not the same. It could be the same for very specific values of the elements of each matrix.