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3 years ago

Factor out the greatest common factor 30c-45d

Mathematics

2 answers:

34kurt3 years ago

8 0

15(2c - 3d) both 30 and 45 are divisible by 15

olga nikolaevna [1]3 years ago

7 0

Simplifying
30c + -45d

Factor out the Greatest Common Factor (GCF), '15'.
15(2c + -3d)

Final result:
15(2c + -3d)

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algol13

Answer:

Substitute n =1, 2, 3, 4, 5 to get the first five terms.

Step-by-step explanation:

(4). $g_n = 2 . 3^{n - 1}$

To find the first term, substitute n = 1.

Therefore, $g_1 = 2 . 3^{1 - 1} = 2 . 3^{0} = 2 . 1$ = 2

$g_2 = 2 . 3^{2 - 1} = 2 . 3$ = 6

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$g_4 = 2 . 3^{4 - 1} = 2 . 3^{3} = 2 . 27$ = 54

$g_5 = 2 . 3^{5 - 1} = 2 . 3^{4} = 2 . 81$ = 168

Therefore the first five terms of the sequence are: 2, 6, 18, 54, 168.

(5). $t_n = \frac{2}{3}t_{n - 1}$

$t_2 = \frac{2}{3} t_{2 - 1} = \frac{2}{3}t_1 = \frac{2}{3}6$ = 4

$t_3 = \frac{2}{3} t_2 = \frac{2}{3}. 4 = \frac{8}{3}$

$t_4 = \frac{2}{3}t_3 = \frac{2}{3}\frac{8}{3} = \frac{16}{9}$

$t_5 = \frac{2}{3}t_4 = \frac{2}{3}\frac{16}{9} = \frac{32}{27}$

Therefore the first five terms in the sequence are: 6, 4, 8/3, 16/9, 32/27.

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