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juin [17]
3 years ago
8

Which region contains the solution

Mathematics
2 answers:
dangina [55]3 years ago
5 0
Hello!

In this case, there are no solutions because the lines are parallel, and the system of inequalities are opposite each other. 


Hope this helps :))
Bas_tet [7]3 years ago
4 0
The correct answer is letter A, There is no solution for the given system of inequalities.
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Hayden has 7 shells . Elly has 5 fewer shells than Hayden . Beth has 2 shells . How many shells do they have in all?
sleet_krkn [62]

Answer:

11 i think

Step-by-step explanation:

5 0
2 years ago
Find the value of x. Show your work, congruent triangles
Nataly_w [17]

Answer:

x = 12

Step-by-step explanation:

130 + y = 180   {linear pair}

          y = 180 - 130

         y = 50°

3x - 6 +8x + 4 + 50 = 180        {Angle sum property of triangle}

3x + 8x - 6 +  4 + 50 = 180          {Combine like terms}

          11x + 48          = 180

                    11x        = 180 - 48

                          11x = 132

                            x = 132/11

x = 12°

5 0
3 years ago
If you have 3 sticks of length 2 cm, 3 cm and 5 cm. What kind of triangle these sticks would form? Justify your answer.
stepan [7]

Answer:

Scalene triangle (As the definition says)

3 0
3 years ago
Please, I need it ASAP!!!! I will give brainliest if correct!!!
LenKa [72]

Answer:

recursive: f(0) = 7; f(n) = f(n-1) -8

explicit: f(n) = 7 -8n

Step-by-step explanation:

The sequence is an arithmetic sequence with first term 7 and common difference -8. Since you're numbering the terms starting with n=0, the generic case will be ...

recursive: f(0) = first term; f(n) = f(n-1) + common difference

explicit: f(n) = first term + n·(common difference)

To get the answer above, fill in the first term and common difference values.

4 0
3 years ago
Write a possible third degree polynomial with integer coefficient that have zeros: 1 2i, -4. Assume the leading coefficient to b
jasenka [17]

Answer:

The polynomial is:

p(x) = x^3 + 2x^2 - 3x + 20

Step-by-step explanation:

A third degree polynomial can be written in function of it's zeros x_1, x_2, x_3 the following way:

p(x) = a(x - x_1)(x - x_2)(x - x_3)

In which a is the leading coefficient.

Integer coefficient that have zeros: 1+2i, 1-2i, -4

Leading coefficient: 1

So

p(x) = 1(x - (1+2i))(x - (1-2i))(x - (-4))

p(x) = (x - 1 -2i)(x - 1 + 2i)(x + 4)

p(x) = ((x-1)^2 - (2i)^2)(x + 4)

p(x) = (x^2 - 2x + 1 - 4i^2)(x + 4)

Since i^2 = -1

p(x) = (x^2 - 2x + 1 + 4)(x + 4)

p(x) = (x^2 - 2x + 5)(x + 4)

p(x) = x^3 + 4x^2 - 2x^2 - 8x + 5x + 20

p(x) = x^3 + 2x^2 - 3x + 20

3 0
3 years ago
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