$\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$

This tells us that:

$A=\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right]$

$b=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$

Step-by-step explanation:

So we are saying we have scalars, c and d, such that:

$c\left[\begin{array}{ccc}5\\5\\ 3\end{array}\right]+d\left[\begin{array}{ccc}7\\-8\\-9\end{array}\right]=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$ .

So we want to find a way to express this as:

Ax=b where x is the scalar vector, $\left[\begin{array}{ccc}c\\d\end{array}\right]$ .

So we can write this as:

$\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$

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What is the distance between the points (–3, 5) and (2, –6)?

DENIUS [597]

the answer is C.12.1 units

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I you were flipping a nickel a dime and a quarter how many possible outcomes are possible for getting heads or tails

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