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otez555 [7]
3 years ago
5

❗️❗️❗️❗️PLEASE HELP MEEE❗️❗️❗️❗️

Mathematics
1 answer:
koban [17]3 years ago
8 0
For this case we have the following inequality:
 - \frac{1}{2}x \ \textgreater \  6
 We multiply both sides of the equation by -1.
 By doing this, change the symbol of the inequality.
 We have then:
 \frac{1}{2}x\ \textless \ -6
 From here, we clear the value of x.
 We have then:
 x \ \textless \ (2) (- 6)

 x \ \textless \ -12
 
 Answer:
 The solution set is:
 
{x | x R, x < -12} 
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2 years ago
Solve the following system
scZoUnD [109]

Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

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Write the set of points from −6−6 to 33 but excluding −2−2 and 33 as a union of intervals:
Softa [21]

In this question, we have to write the set of points from -6 to 3 but excluding -2 and 3 as a union of intervals .

For union, we use U.

For the points that we have to exclude , we put parenthesis on there side  that is  ().

Therefore the required interval form is

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As we see that () are used with -2 and 3 . And that's the required interval form .

5 0
3 years ago
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