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2 years ago

How would you put this in a graph

Mathematics

1 answer:

Lerok [7]2 years ago

4 0

Answer:

I would suggest using a app called desmos if you put in the table it will make the graph for you :)

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What is the answer??

Svetradugi [14.3K]

Answer:

32.

Step-by-step explanation:

I accidentally put my age too high. Im only in middle school but 32 could be one answer. If two sides are the same then it could be an isoceles triangle or whatever.

6 0

2 years ago

Explain how drawing triangles can determine the sum of the interior angles of the quadrilateral?

Kobotan [32]

Answer: each triangle drawing is 180 degrees total, 2 triangles is 360 etc.

Step-by-step explanation:

3 0

2 years ago

*THIS IS FROM K12 MIDDLE SCHOOL*

AlexFokin [52]

Answer:

Step-by-step explanation:

the answer is 7 because -4 in the first pair and 3 from the second ordered pair is 7.The other side is 4 because 2 and -2 are 4 places away from each other

hope it helps :)

7 0

2 years ago

Read 2 more answers

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

2 years ago

Eli is moving to a new city. He packs his belongings in cube-shaped boxes that each have a volume of one cubic foot. He places t

klemol [59]

Answer:

jnnm mnn m

Step-by-step explanation:

4 0

2 years ago