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3 years ago

Can anyone please help me with this? I haven't went to classes and I need this answered real quick. Anyone know it?

Mathematics

2 answers:

Vera_Pavlovna [14]3 years ago

7 0

y=mx+b

7 = 5m + 10

subtract 10 from both sides

7 - 10 = 5m + 10 - 10

-3 = 5m

divide both sides by 5

-3/5 = 5m/5

-3/5 = m

your answer is -3/5

sesenic [268]3 years ago

6 0

In this case, you are just going to substitute the given values for the variables and solve:

$7 = 5m + 10$

$-3 = 5m$

$m = -\dfrac{3}{5}$

The answer would be Choice C, The slope is -3/5.

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Answer:

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3 years ago

If sin α=(12/13), and cos α=(5/13), then tan α = ?

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3 years ago

RS = 4x – 9, ST = 19, RT = 80 – 14 Equation: 82 – 14 RS RT

algol13

Step-by-step explanation:

The expressions are not properly written

Given

RS = 4x – 9

ST = 19

RT = 8x – 14

Based on the given parameters, the addition postulate below is true

RS+ST = RT

Substitute

4x-9+19 = 8x-14

collect like terms

4x-8x = -14-19+9

-4x = -33+9

-4x = -24

x = -24/-4

x = 6

Get RS:

RS = 4x-9

RS = 4(6)- 9

RS = 24-9

RS = 15

Get RT:

RT = 8x - 14

RT = 8(6)-14

RT = 48-14

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4 0

3 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago