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2 years ago

Can someone please help ASAP!! Pleaseee I’ll give brainliest!!!

Mathematics

2 answers:

Grace [21]2 years ago

6 0

Answer:

all real numbers

Step-by-step explanation:

The graphs of positive and negative x^2 parabolas will always have a domain of all real numbers. Even though you only have a portion of the graph and see a "restriction" on your domain values, it is incorrect to assume that the domain is limited to what you can see. As the branches of the parabola keep going up and up and up, the values of x keep getting bigger and bigger and bigger. Again, this is true for all + or - parabolas. ....... Can consider brainlist*:)

prohojiy [21]2 years ago

3 0

I believe that it is All Real Numbers because it is asking for the domain of function f, which is referring to the x axis only. If you were to graph that as an inequality, you’d come to the conclusion that the line could extend to all possible numbers on the x-axis.

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Inscribed angles equal half of their arc length. So <D+ <B = 180
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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.

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2 years ago