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Mkey [24]
1 year ago
15

Evaluate the expression for s = -15. -S= ​

Mathematics
1 answer:
Oxana [17]1 year ago
3 0

Answer:

  -S = 15

Step-by-step explanation:

You are asked to find the opposite of the given value.

<h3>Opposite</h3>

The opposite of an integer value is obtained by changing its sign.

  S = -15

  -S = -(-15) = 15

__

<em>Additional comment</em>

Equivalently, you are multiplying by -1. The product of an even number of minus signs is a plus sign. We usually don't write the plus sign for a numerical value.

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3 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
2 years ago
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