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polet [3.4K]
3 years ago
15

Write an inequality and show on a number line all numbers: from (–3) to 2 exclusives.

Mathematics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

The inequality is -3 < x < 2

The number line is included

Step-by-step explanation:

To express the numbers as an inequality, we have for exclusive numbers the form, [-3, 2] which gives;

-3 < x < 2

We note that an exclusive number (or larger than and lesser than inequality) is represented by an empty in the numbers excluded (for strict equality)

Therefore, we draw a number line with empty circles at number points -3, and 2 as follows;

-4 -3 -2 -1 0 2 1 2 3 4 5

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Rewrite the equation in vertex form. then find the vertex of the graph. y=-3x^2-5x+1
amm1812

Answer: y = -3(x + \frac{5}{6})² + \frac{37}{12}, (-\frac{5}{6}, \frac{37}{12})

<u>Step-by-step explanation:</u>

First, you need to complete the square:

y   = -3x² - 5x + 1

<u> -1  </u>   <u>                -1  </u>

y - 1 = -3x² - 5x

y - 1 = -3(x² + \frac{5}{3}x

y - 1 + -3(\frac{25}{36}) = -3(x² + \frac{5}{3}x + \frac{25}{36})

           ↑                     ↓            ↑

                                  \frac{5}{3*2} = (\frac{5}{3*2})^{2}

y - 1 - \frac{25}{12} = -3(x + \frac{5}{6})²

y - \frac{12}{12} - \frac{25}{12} = -3(x + \frac{5}{6})²

y  - \frac{37}{12} = -3(x + \frac{5}{6})²

y = -3(x + \frac{5}{6})² + \frac{37}{12}

Now, it is in the form of y = a(x - h)² + k   <em>where (h, k) is the vertex</em>

Vertex = (-\frac{5}{6}, \frac{37}{12})

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