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pantera1 [17]
2 years ago
15

Shaunta is developing a recursive formula to represent an arithmetic sequence in which 5 is added to each term to determine each

successive term. Which formula could represent their sequence? 0.36, 0.26, 0.16, 0.06, -0.04, -0.14… — -0.1 — -0.01 — 0.01 — 0.1
Mathematics
1 answer:
Fantom [35]2 years ago
3 0

Answer:

f(n+1) = f(n) + 5

Step-by-step explanation:

Required

Recursive function to represent the scenario

Let the current term be f(n);

This means that the next term will be: f(n + 1);

From the question, 5 is added to f(n) to get f(n + 1).

Hence, the function is:

f(n+1) = f(n) + 5

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Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2 ( using the property (ab)^m=a^m.b^m )

=(2^6)(3^{-12})\times (2^{-6})(3^4) ( using the property (a^m)^n=a^{mn}

=(2^6)(2^{-6})(3^{-12})(3^4) ( combining the like powers )

=2^{6-6}3^{-12+4} ( using the property a^m.a^n=a^{m+n} )

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Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

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