9514 1404 393
Answer:
A. 3×3
B. [0, 1, 5]
C. (rows, columns) = (# equations, # variables) for matrix A; vector x remains unchanged; vector b has a row for each equation.
Step-by-step explanation:
A. The matrix A has a row for each equation and a column for each variable. The entries in each column of a given row are the coefficients of the corresponding variable in the equation the row represents. If the variable is missing, its coefficient is zero.
This system of equations has 3 equations in 3 variables, so matrix A has dimensions ...
A dimensions = (rows, columns) = (# equations, # variables) = (3, 3)
Matrix A is 3×3.
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B. The second row of A represents the second equation:

The coefficients of the variables are 0, 1, 5. These are the entries in row 2 of matrix A.
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C. As stated in part A, the size of matrix A will match the number of equations and variables in the system. If the number of variables remains the same, the number of rows of A (and b) will reflect the number of equations. (The number of columns of A (and rows of x) will reflect the number of variables.)
When you have two unit prices for two comparable items, you have to convert one of the prices to the other one.
Answer:
In this equation x is equal to -5.
Step-by-step explanation:
In order to find this, we first have to note for the base (2) to be larger than the answer (1/32), the exponent would have to be negative. We also can note that 2^5 is equal to 32, which allows us to know that the negated version would give us 1/32.
2^-5 = 1/32
5/11 because if you simplify starting at 40/88 you get 5/11