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irakobra [83]
3 years ago
8

How to solve 3y² + 2y - 8​

Mathematics
2 answers:
Ber [7]3 years ago
6 0
(Y+2) (3y-4) I hope that it’s right
Lyrx [107]3 years ago
3 0

Answer:

(y + 2) (3y - 4)

Step-by-step explanation:

1. 3y^2 + 2y - 8

Rewrite the expression.

2. 3y^2 + 6y - 4y - 8

Factor the expression.

3. 3y(y + 2) - 4(y + 2)

Factor the expression once again.

4. (y + 2) (3y - 4)

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Help.? geometry basics homework 2: segment addition postulate
Neporo4naja [7]

The value of x is 3 given that ST = 8x + 11 and TU = 12x-1

The point that bisects a line divides the line into two equal parts

If T is the midpoint of SU, the following are true:

  • ST = TU
  • ST + TU = SU

Given the following

ST = 8x + 11

TU = 12x-1

Since ST = TU

8x+11 = 12x-1

8x - 12x = -1 - 11

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x = 12/4

x = 3

Hence the value of x is 3 given that ST = 8x + 11 and TU = 12x-1

Learn more here: brainly.com/question/17204733

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Use the discriminant to describe the roots of each equation. Then select the best description.
Paha777 [63]

The <em>quadratic</em> equation 3 · x² + 7 · x - 2 = 0 has a <em>positive</em> discriminant. Thus, the expression has two <em>distinct real</em> roots (<em>real</em> and <em>irrational</em> roots).

<h3>How to determine the characteristics of the roots of a quadratic equation by discriminant</h3>

Herein we have a <em>quadratic</em> equation of the form a · x² + b · x + c = 0, whose discriminant is:

d = b² - 4 · a · c     (1)

There are three possibilities:

  1. d < 0 - <em>conjugated complex</em> roots.
  2. d = 0 - <em>equal real</em> roots (real and rational root).
  3. d > 0 - <em>different real</em> roots (real and irrational root).

If we know that a = 3, b = 7 and c = - 2, then the discriminant is:

d = 7² - 4 · (3) · (- 2)

d = 49 + 24

d = 73

The <em>quadratic</em> equation 3 · x² + 7 · x - 2 = 0 has a <em>positive</em> discriminant. Thus, the expression has two <em>distinct real</em> roots (<em>real</em> and <em>irrational</em> roots).

To learn more on quadratic equations: brainly.com/question/2263981

#SPJ1

7 0
2 years ago
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