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statuscvo [17]
3 years ago
12

For the graph x = 1 find the slope of a line that is perpendicular to it and the slope of a line parallel to it. Explain your an

swer with two or more sentences.
Mathematics
1 answer:
scoray [572]3 years ago
5 0
Lines that are perpendicular to another line has an opposite reciprocal slope. (opposite meaning to negate the number).
Lines that are parallel have the same slope.

But the line you provided is a vertical line, and vertical lines are undefined. There is no possible perpendicular or parallel line. Maybe you gave the wrong equation and didn't put it in <em>y=mx+b</em> form. If so, you can correct it in the comments, and if not, that's you answer!

:)
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Please help if you can pic attatched
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]

3 0
3 years ago
Three students are trying to find the diameter of circle C. Gregory believes the diameter is equal to the length of chord XY. Ma
kotegsom [21]

Answer: I know the answer is 75 units but I’m not sure who’s right

Step-by-step explanation:

4 0
3 years ago
NEED HELP ASAP
Jobisdone [24]

ANSWER: A. 46

 

 

SOLUTION

 

Given that Q is equidistant from the sides of TSR

m∠TSQ = m ∠QSR

 

To solve for x

m∠TSQ = 3x + 2

m ∠QSR = 8x – 33

 

Since m∠TSQ = m ∠QSR

3x + 2 = 8x – 33

 

Add 33 to both sides

3x + 2 + 33 = 8x – 33 + 33

3x + 35 = 8x

8x = 3x + 35

 

Subtract 3x from both sides

8x – 3x = 3x – 3x + 35

5x = 35

 

Divide both sides by 5

x = 7

 

Since m∠TSQ = 3x + 2, and x = 7

m∠TSQ = (3*7) + 2

m∠TSQ = 21 + 2

m∠TSQ = 23

 

To solve for RST

Given that Q is equidistant from the sides of RST

m∠RST = m∠TSQ + m ∠QSR

Since m∠TSQ = m ∠QSR

m∠RST = 2m∠TSQ = 2m ∠QSR

 

Ginen, m∠RST = 2m∠TSQ

m∠TSQ = 23

m∠RST = 2(23)

m∠RST = 46

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WARRIOR [948]

Answer:

Step-by-step explanation:

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notka56 [123]
3= (-2+X)/2
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3=(1.5+Y)/2
6=1.5+Y
4.5=Y

G=(8,4.5)
5 0
3 years ago
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