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Mathematics

1 answer:

SashulF [63]3 years ago

7 0

Answer:

A.7

B.4

Step-by-step explanation:

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Find the slope of the line passing through the given points.

deff fn [24]

Answer:

<h3>The answer is - 3</h3>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

From the question we have

$m = \frac{8 - 5}{ - 8 - - 7} = \frac{3}{ - 8 +7 } = - \frac{3}{1} = - 3 \\$

We have the final answer as

Hope this helps you

5 0

2 years ago

1.5 > 0.75x + 1 <br> what is x

motikmotik

Answer:

2/3 > x

Step-by-step explanation:

0.5 > 0.75x

2/3 > x

6 0

2 years ago

Solve the following matrix equations: (matrices)

Masja [62]

Step-by-step explanation:

$3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\ \\ 3X = \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \frac{1}{3} \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \begin{pmatrix} \frac{ - 3}{3} & \frac{3}{3} \\ \\ \frac{6}{3} & \frac{9}{3} \end{pmatrix}\\ \\ \huge \red{ X} = \purple{ \begin{pmatrix} - 1 &1 \\ 2 & 3 \end{pmatrix}}$

$3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\begin{pmatrix} \frac{3}{3} & \frac{0}{3} & \frac{-3}{3} \\\\ \frac{6}{3} & \frac{3}{3} & \frac{0}{3} \\\\ \frac{9}{3} & \frac{6}{3} & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1 & 0 & - 1\\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}}\\$