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frutty [35]
3 years ago
15

There are 3 different mathematics courses, 3 different science courses, and 5 different history courses. If a student must take

one of each, how many different ways can this be done?
Mathematics
1 answer:
Damm [24]3 years ago
3 0

Answer:

45 ways

Step-by-step explanation:

We are given;

there are 3 different math courses, 3 different science courses, and 5 different history courses.

Thus;

Number ways to take math course = 3

The number of ways to take science course = 3

The number of ways to take history course = 5

Now, if a student must take one of each course, the different ways it can be done is;

possible ways = 3 x 3 x 5 = 45 ways.

Thus, number of different ways in which a student must take one of each subject is 45 ways.

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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

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