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Olegator [25]
3 years ago
5

20 POINTS PLUS A BRAINLIEST IF ITS RIGHT!!!

Mathematics
1 answer:
kaheart [24]3 years ago
4 0
I believe answer is any real number
so

<span>D.
x can be any real number</span>
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Plz help very important
allsm [11]
I hate questions like this. It's like wishing for more wishes.
7 0
3 years ago
How do I find the area of an octagon which has 7m on each side and the height is 8.8m
Naddika [18.5K]

Answer:

236.59

Step-by-step explanation:

The formula 2(1+\sqrt{2}) l^{2} can be used. Substitute the length of one side into the equation and solve.

5 0
2 years ago
Match each equation on the left with the number and type of its solutions on the right.
klemol [59]

Answer:

Step-by-step explanation:

1). Given equation is,

   2x² - 3x = 6

   2x² - 3x - 6 = 0

   To find the solutions of the equation we will use quadratic formula,

   x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

   Substitute the values of a, b and c in the formula,

   a = 2, b = -3 and c = -6

   x = \frac{3\pm\sqrt{(-3)^2-4(2)(-6)}}{2(2)}

   x = \frac{3\pm\sqrt{9+48}}{4}

   x = \frac{3\pm\sqrt{57}}{4}

   x = \frac{3+\sqrt{57}}{4},\frac{3-\sqrt{57}}{4}

   Therefore, there are two real solutions.

2). Given equation is,

    x² + 1 = 2x

    x² - 2x + 1 = 0

    (x - 1)² = 0

     x = 1

     Therefore, there is one real solution of the equation.

3). 2x² + 3x + 2 = 0

     By applying quadratic formula,

     x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

      x = \frac{-3\pm\sqrt{3^2-4(2)(2)}}{2(2)}

      x = \frac{-3\pm\sqrt{9-16}}{4}

      x = \frac{-3\pm i\sqrt{7}}{4}

      x = \frac{-3+ i\sqrt{7}}{4},\frac{-3- i\sqrt{7}}{4}

      Therefore, there are two complex (non real) solutions.

3 0
2 years ago
I need help on this please SHOW WORK
elena-s [515]

\boxed{\large{\bold{\blue{ANSWER~:) }}}}

See this attachment

7 0
3 years ago
Marika solved the equation (6x + 15)2 + 24 = 0. Her work is below. 1. (6x + 15)2 = –24 2. StartRoot (6 + 15) squared EndRoot = S
malfutka [58]

Answer:

There are no real solutions to this equation because the square root of a negative number is not real. So answer B

Step-by-step explanation:

11 0
3 years ago
Read 2 more answers
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