3 years ago

Please answer the question from the attachment.

1 answer:

6 0

$\frac{y^{2} - 7y + 12}{2y^{2} - 9y + 9} = \frac{y^{2} - 4y - 3y + 12}{2y^{2} - 6y - 3y + 9} = \frac{y(y) - y(4) - 3(y) - 3(-4)}{2y(y) - 2y(3) - 3(y) - 3(-3)} = \frac{y(y - 4) - 3(y - 4)}{2y(y - 3) - 3(y - 3)} = \frac{(y - 3)(y - 4)}{(2y - 3)(y - 3)} = \frac{y - 4}{2y - 3}$

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C(n) = 4/9(-3)^n-1 What is the 3rd term in the sequence?

Yanka [14]

Step-by-step explanation:

$c(n) = \frac{4}{9} ( - 3)^{n - 1} \\ \\ \therefore \: c(3) = \frac{4}{9} ( - 3)^{3- 1} \\ \\ \therefore \: c(3) = \frac{4}{9} ( - 3)^{2} \\ \\ \therefore \: c(3) = \frac{4}{9} \times 9 \\ \\ \huge{ \purple{ \boxed{\therefore \: c(3) = 4}}}$