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Wewaii [24]
3 years ago
11

Help please! Will award brainliest!

Mathematics
1 answer:
emmasim [6.3K]3 years ago
5 0

C. (2x - 5)

<u>Explanation:</u>

Given:

The factors of 4x^2 - 25 = ?

The binomial factor can be simplified and written as

4x^2 - 25 = (2x + 5) (2x - 5)

When these two factors are multiplied then they give 4x² - 25

Thus, (2x + 5) and (2x - 5) are the two factors of 4x² - 25

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Debora [2.8K]
It means that all values of x would make it true. Also take a look at the equation when we simplify:

4x + 7 = 1 + 4x + 6
Which simplified turns into
4x + 7 = 4x + 7

So no matter what value you put for x, it will be the same on both sides, making all values true for x.
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3 years ago
Use the table to find the slope.<br><br><br> -2<br><br><br> 2<br><br><br> 1/2<br><br><br> -1/2
Tresset [83]

Answer:

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Step-by-step explanation:

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3 years ago
Find 2 a for a = 3 1/4<br> 1) 6 1/2<br> 2) 6 1/4<br> 3) 23 1/4<br> 4) 5 1/2
Nataly_w [17]

Answer:

  1)  6 1/2

Step-by-step explanation:

Put 3 1/4 where "a" is and do the arithmetic.

  2(3 1/4) = 2·3 + 2·(1/4) = 6 2/4 = 6 1/2

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3 years ago
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Elanso [62]

The series of operations for each case are listed below:

  1. GCF / GCF / GCF
  2. GCF / Grouping
  3. Quadratic trinomial
  4. GCF / Quadratic trinomial
  5. Difference of squares
  6. Difference of cubes / Quadratic trinomial
  7. Sum of cubes
  8. GCF / Quadratic trinomial
  9. GCF / Difference of squares

<h3>How to applying factor properties to simplify algebraic expressions</h3>

In algebra, factor properties are commonly used to solve certain forms of polynomials in a quick and efficient way and whose effectiveness is sustained on all definitions and theorems known in real algebra. In this problem, we should explain and show what factor properties are used in each case:

Case 1

5 · x · y³ + 10 · x² · y                                             Given

5 · (x · y³ + 2 · x² · y)                                            GCF

5 · x · (y³ + 2 · x · y)                                              GCF

5 · x · y · (y² + 2 · x)                                              GCF

Case 2

6 · z · x + 9 · x + 14 · z + 21                                   Given

3 · x · (z + 3) + 7 · (z + 3)                                       GCF

(3 · x + 7) · (z + 3)                                                  Grouping

Case 3

a² + 2 · a - 63                                                       Given

(a + 9) · (a - 7)                                                       Quadratic trinomial

Case 4

6 · z² + 5 · z - 4                                                     Given

6 · [z² + (5 / 6) · z - 2 / 3]                                      GCF

6 · (z - 1 / 2) · (z + 4 / 3)                                         Quadratic trinomial

Case 5

81 · m² - 25                                                           Given

(9 · m + 5) · (9 · m - 5)                                           Difference of squares

Case 6

8 · x³ - 27                                                               Given

(2 · x - 3) · (4 · x² + 6 · x + 9)                                  Difference of cubes

4 · (2 · x - 3) · [x² + (3 / 2) · x + 9 / 4]                      Quadratic trinomial

Case 7

27 · b³ + 64 · z³                                                      Given

(3 · b + 4 · z) · (9 · b² - 12 · b · z + 16 · z²)               Sum of cubes

Case 8

2 · w³ - 28 · w² + 80 · w                                         Given

2 · w · (w² - 14 · w + 40)                                          GCF

2 · w · (w - 4) · (w - 10)                                             Quadratic trinomial

Case 9

200 · a⁴ - 18 · b⁶                                                     Given

2 · (100 · a⁴ - 9 · b⁶)                                                GCF

2 · (10 · a² + 3 · b³) · (10 · a² - 3 · b³)                       Difference of squares

To learn more on polynomials: brainly.com/question/17822016

#SPJ1

7 0
1 year ago
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