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.)
Answer:
(- 1/2 , √3/2)
Step-by-step explanation:
t=20π/3
t' = 20π/3 - 6π = 2π/3 (120°) ... 2nd quadrant
If we start from (1,0) of unit circle, the coordinate of terminal point (x,y)
OF/OE = - cos 60° = x / 1 = - 1/2
FE/OE = sin 60° = y/1 = √3/2
Answer:
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1
Set X: {51,49,37,28,33,39,49,42} <br><br>
Set Y :{23,2934,36,39,22,25,33}
SpyIntel [72]
Answer:
X
Step-by-step explanation:
126
Answer:
30
Step-by-step explanation:
There are 6 ways the president can be chosen. Assuming the vice-president must be a different person, there are then 5 possibilities for that office. The total number of possible choices is ...
6 × 5 = 30
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Each of the 6 president choices can have any of 5 different vice-president choices.