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3 years ago

What is the 6th term in the sequence described by the following recursive formula?

Mathematics

1 answer:

larisa86 [58]3 years ago

4 0

Answer:

Correct answer is: D) a₆ = - 11

Step-by-step explanation:

Given:

a₁ = 9 and aₙ = aₙ₋₁ - 4

We know it is

a₆ = a₅ + d also it is given a₆ = a₅ - 4

When we equate the right sides of equality we get:

a₅ + d = a₅ - 4 => d = - 4

We also know it is a₆ = a₁ + 5 d

Now we will find a₆

a₆ = 9 + 5 (-4) = 9 - 20 = - 11

a₆ = - 11

God with you!!!

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3 years ago

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2 years ago

Read 2 more answers

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

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2 years ago

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Answer:

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3 years ago

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vovikov84 [41]

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