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saveliy_v [14]
3 years ago
14

Write a recursive definition for the sequence 14, 10, 6, 2,… Common difference is 4 what do I do next?

Mathematics
2 answers:
Reil [10]3 years ago
8 0
F(x) = f(x-1) - 4, with f(0) = 14  
Since you're looking for a recursive definition, you simply need to do 2 things.
 1. Define the function for some constant starting point(s).
 2. Define the function in terms of itself for all other points. 
 For this question, I'll assume the first point is f(0). So we have f(x) = ?, with f(0) = 14 
 And we've handed the starting point by simply declaring that f(0) has the value of 14. Now for the recursive portion.
 As you've noticed, each subsequent value is 4 less than the previous. So we can say f(x) = f(x-1) - 4. And with that, we can make the function.
 f(x) = f(x-1) - 4, with f(0) = 14 
 As a side note, you are not restricted to just 1 starting value, nor just the immediate prior invocation of the function. For instance the definition of the Fibonacci sequence could be defined as
 f(x) = f(x-1) + f(x-2), with f(0) = 0 and f(1) = 1
leonid [27]3 years ago
6 0
A[1] = 14 . . . . . . . . . defines the first term
a[n+1] = a[n] -4 . . . says each term is 4 less than the one before it
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2 years ago
Simplify (-2x^4y^-3)^4
elixir [45]

Answer:

( - 2 {x}^{4} {y}^{ - 3} )^{4}  =  \frac{16 {x}^{16}}{ {y}^{12} }

Step-by-step explanation:

We want to simplify:

( - 2 {x}^{4} {y}^{ - 3} )^{4}

We need to apply the exponential property for products of powers.

Recall that:

( {a}^{m}  \times  {a}^{n} )^{p}  =  {a}^{mp}  \times  {a}^{np}

We apply this rule to get:

( - 2 {x}^{4} {y}^{ - 3} )^{4}  = ( { - 2)}^{4} x \times {x}^{16}  \times {y}^{ - 12}

This simplifies to:

( - 2 {x}^{4} {y}^{ - 3} )^{4}  = 16 {x}^{16}  \times {y}^{ - 12}

We rewrite as positive index to get:

( - 2 {x}^{4} {y}^{ - 3} )^{4}  =  \frac{16 {x}^{16}}{ {y}^{12} }

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3 years ago
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sammy [17]
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4 0
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Gnesinka [82]
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