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2 years ago

-6x+6y=9 (1/2,2) please helppppp

Mathematics

1 answer:

morpeh [17]2 years ago

7 0

Answer:

Not a solution.

Step-by-step explanation:

$-6x + 6y = - 9\\\rule{150}{0.5}\\-6(\frac{1}{2}) + 6(2) = -9\\\\-3 + 12 = -9\\\\9 =-9$

The final result is a contradiction. It is not a solution to the equation, and therefore, will not be on the line.

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Marissa has a photograph that measures 2 in. by 4 in. She has mounted the picture on a mat so that there is a border that measur

DanielleElmas [232]

Step-by-step explanation:

The initial image of the photo is 2 in by 4 in. The mat is 4 in by 6 in.

The new image is dilated by a scale of 2. So we double the dimensions. The new photo is 4 in by 8 in. The new mat is 8 in by 12 in.

3 0

3 years ago

How to solve 1/4 - -4/3 - -1 2/9

Ahat [919]

Answer:

2 29/36

Step-by-step explanation:

So circle the two signs., so the problem is now this:

1/4+4/3+1 2/9

Then you find the GCM, or Greatest Common Denominator.

It is 36. Then, you re-write the new answers

9/36+ 48/36+ 1 18/36

Now we have this:

1 65/36

The answer is 2 29/36

8 0

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

What is the answer to -5+2=

Pavlova-9 [17]

Answer:

-3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The owner of a football team claims that the average attendance at games is over 523, and he is therefore justified in moving th

USPshnik [31]

Answer:

C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.

Step-by-step explanation:

Let μ be the the average attendance at games of the football team

The claim: the average attendance at games is over 523

Null and alternative hypotheses are:

$H_{0}$ : μ=523

$H_{a}$ : μ>523

The conclusion is failure to reject the null hypothesis.

This means that test statistic is lower than critical value. Therefore it is not significant, there is no significant evidence to accept the alternative hypothesis.

That is no significant evidence that the average attendance at games of the football team is greater than 523.

7 0

2 years ago