The 9th graders are selling tickets to raise
1 answer:
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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:
Option B. t=5/3w ;1/3
Step-by-step explanation:
We are told that varies Inversely with
thus generally speaking
∝
since they are inverted, <u>otherwise</u> if they were proportionally varied then
∝
. This means that there is a constant value (
) for which
is inversely proportional to
and can be mathematically expressed as:
Eqn. (1)
Now since we are given the values of and
, we can plug them in Eqn. (1) and find our constant of proportionality
as follow:
Now that we have our constant we can find the new value for the second value of
as follow:
Therefore based on the options give, we can see that Option B. is correct since
and for
, then