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Lerok [7]
2 years ago
11

Which recursive formula can be used to generate the sequence below, where f(1) = 6 and n ≥ 1? 6, 1, –4, –9, –14, … f (n 1) = f(n

) 5 f (n 1) = f(n) – 5 f (n) = f(n 1 ) – 5 f (n 1) = –5f(n)
Mathematics
1 answer:
marysya [2.9K]2 years ago
7 0

The recursive formula is f(n+1)= f(n)-5.

<h3>What is recursive function?</h3>

Given sequence :

f(1)= 6, n≥1.

Let f(n) denote the nth term of the given sequence.

f(1)= 6

f(2)= 1 = f(1)-5

f(3)= -4 = 1-5= f(2) -5

f(4)= -9 = 1-10 = f(2)-10

and so on

Hence, the recursive formula is f(n+1)= f(n)-5.

Learn more about this concept here:

brainly.com/question/13210754

#SPJ1

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If f(x) = - x + 4 and g(x) = x ^ 2 - 5 , then f(g(-2))=?
kkurt [141]

Answer:

Step-by-step explanation:

using your f(x) I want to show you how to plug in different things itnot he function.

f(x) = -x + 4

f(1) = -1 + 4 = 3

f(2) = -2 + 4 = 2

f(m) = -m + 4

f(abc) = -(abc) + 4

f(h(x)) = -h(x) + 4

Does that help?  what if you replaced h(x) with g(x)?  Of course you already have that x so you could make it a little simpler as well.  Let me know if you don't quite get it.

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Which expression is equivalent to...? Screenshots attached. Please help! Thank you.
Studentka2010 [4]

Answer:

4x^{3} y^{2} (\sqrt[3]{4 x y})

Step-by-step explanation:

Another complex expression, let's simplify it step by step...

We'll start by re-writing 256 as 4^4

\sqrt[3]{256 x^{10} y^{7} } = \sqrt[3]{4^{4} x^{10} y^{7} }

Then we'll extract the 4 from the cubic root.  We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.

\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }

Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.

4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }

For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.

4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}

The answer is then:

4x^{3} y^{2} \sqrt[3]{4 x y} = 4x^{3} y^{2} (\sqrt[3]{4 x y})

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