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goldfiish [28.3K]
3 years ago
9

The solution set of the inequality 2x − y > 2 consists of all the points __ ? the line __ ?

Mathematics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

  • b - below; y = 2x − 2

Step-by-step explanation:

  • 2x − y > 2
  • 2x - y + y - 2 > 2 + y - 2
  • 2x - 2 > y
  • y < 2x - 2

Solution is the points below the line y = 2x - 2

<u>Correct option is:</u>

  • b - below; y = 2x − 2
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Math Questions :3 thank uu &lt;3
coldgirl [10]
14.       1.5, 10    <- Answer
15.       5,1          <- Answer

Proof 14 

Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5               
                y = 10

Proof 15. 

Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer:  {x = 5                           y = 1
8 0
3 years ago
A number increased by the sum of the number and four is 20. Find the number.
miss Akunina [59]

Answer:

i think 5 is the right answer

7 0
3 years ago
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vesna_86 [32]

Answer: 1st one

Step-by-step explanation:

8 0
3 years ago
A square with sides of 3 squareroot 2 is inscribed in a circle. what is the area of one of the sectors formed by the radii to th
mrs_skeptik [129]
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2).  Draw this to verify this statement.  Note that the height of each such triangular area is (3 sqrt(2))/2.

So now we have the base and height of one of the triangular sections.

The area of a triangle is A = (1/2) (base) (height).  Subst. the values discussed above,   A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2).  Show that this boils down to A = 9/2.

You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section.   Doing the problem this way, we get (1/4) (3 sqrt(2) )^2.  Thus, 
A = (1/4) (9 * 2) = (9/2).  Same answer as before. 
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3 years ago
Using only the digits 0 -9 enter a postive vaklue for x
Ghella [55]

Answer:

math w a y .com

Step-by-step explanation:

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