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2 years ago

12

Which point is on the graph of f(x) = 6x? A. (0, 6) B. (1, 6) C. (6, 1) D. (0, 0)

2 answers:

andreev551 [17]2 years ago

6 0

Answer:

D. (0,0)

Step-by-step explanation:

F(x)=6x

0=6x

x=0

Effectus [21]2 years ago

5 0

Answer:

B and D

Step-by-step explanation:

$f(x)=6x$

$y=f(x)$

Let's test all the alternatives given:

$A. (0, 6)$

$f(0)=6 \cdot 0$

$f(0)=0$

$y=0$

Point: $(0, 0)$

This is not the answer

$B. (1, 6)$

$f(1)=6 \cdot 1$

$f(1)=6$

$y=6$

Point: $(1, 6)$

This may be the answer

$C. (6, 1)$

$f(6)=6 \cdot 6$

$f(6)=36$

$y=36$

Point: $(6, 36)$

This is not the answer

$D. (0, 0)$

$f(0)=6 \cdot 0$

$f(0)=0$

$y=0$

Point: $(0, 0)$

This may be the answer

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3 years ago

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

3 years ago