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baherus [9]
2 years ago
12

What is the slope of the line given below?

Mathematics
1 answer:
Masteriza [31]2 years ago
8 0

Answer:

slope = -2

Step-by-step explanation:

You can solve this several different ways, but I’ll show you using rise over run. As you can see, you need to rise 4 to get to the point (0,4) from the point (2,0). Then, you go two units to the left getting 4 as the rise and -2 as the run. Now you divide 4 by -2 to get -2 as the slope.

Hope that helps and please let me know if you need more help!

Have an amazing day!!

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8) 18, 27, 27, 58

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8 0
3 years ago
Find the range of this data set: 225 342 288 552 263.
Katarina [22]
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4 0
3 years ago
Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).
ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

\overrightarrow {PQ}==

\overrightarrow {PR}==

The nonzero vector orthogonal to the plane through the points P,Q, and R is

\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}

\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}

Expand along row 1.

\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}

\overrightarrow n=i(-9)-j(8)+k(2)

\overrightarrow n=-9i-8j+2k

\overrightarrow n=

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

8 0
3 years ago
Please help me asap! Thank you!
blsea [12.9K]
......bruh okay......
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3 years ago
−px+r=−8x−2<br> solve for X
Arlecino [84]
-8x-2=0
-8x=2
8x=-2
x=-1/4
5 0
3 years ago
Read 2 more answers
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