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slamgirl [31]
3 years ago
9

What is the slope of the line that passes through the points (-3,1) and (7,-14)

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

-  \frac{3}{2}

Step-by-step explanation:

\boxed{gradient =  \frac{y1 - y2}{x1 - x2} }

Using the formula above,

slope of line

=  \frac{1 - ( - 14)}{ - 3 - 7}  \\  =  \frac{1 + 14}{ - 10}  \\  =  \frac{15}{ - 10}  \\  =  -  \frac{3}{2}

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asambeis [7]
The answer is B. <span>$78.20; this is the amount Derek pays for 320 minutes of international calls.

Simply substitute in 320 wherever there is an x and solve. The problem also already told us that x stands for minutes so we know it is 320 minutes.</span>
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3 years ago
Could a set of three vectors in ℝ^4 span all ℝ^4? Explain. What about n vectors in ℝ^m when n is less than m?
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Answer:

Check the ecplanation

Step-by-step explanation:

A set of three vectors in R^{4} represents a matrix of 3 column vectors, and each vector containing 4 entries (that is, a matrix of 4 rows, and 3 columns).

Let A be that 4x 3 matrix. The columns of A span R^{m}. if and only if A has a pivot position in each row. So, there are at most 3 pivot positions in the matrix A, but the number of rows is 4, therefore, there exist at least one row not having a pivot position. If A does not have a pivot position in at least one row, then the columns of A do not span R^{4}. It implies that the set of 3 vectors of A does not span all of R^{4}.

In general, the set of n vectors in R^{m} represents a matrix of in rows, and n columns (an in x matrix). So, there are at most n pivot positions in the matrix A, but n is less than the number of rows. In therefore, there exist at least one row that does not contain a pivot position.

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8 0
3 years ago
Walt is mixing fruit punch for a party. He combines 1 gallon 2 quarts 3 pints of orange juice, 1 gallon 3 quarts 7 pints of pine
mafiozo [28]
1 gal + 2 qt + 3 pts
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1 gal + 3 qt + 3 pts
----------------------------add
3 gal + 8 qt + 13 pts

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1 gal = 8 pts
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5 0
3 years ago
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Answer:

v = 16.

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3 years ago
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