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Free_Kalibri [48]
3 years ago
15

Help me please I need good grades for a phone!!! <3

Mathematics
1 answer:
irga5000 [103]3 years ago
8 0

Answer:

Can you show the full problem so that I can help

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At a middle school, 30% of the
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Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.
-BARSIC- [3]

Answer:

Step-by-step explanation:

Hello, please consider the following.

3\cdot 4^n+51=3\cdot 4^n+3\cdot 17=3(4^n+17)

So this is divisible by 3.

Now, to prove that this is divisible by 9 = 3*3 we need to prove that

4^n+17 is divisible by 3. We will prove it by induction.

Step 1 - for n = 1

4+17=21= 3*7 this is true

Step 2 - we assume this is true for k so 4^k+17 is divisible by 3

and we check what happens for k+1

4^{k+1}+17=4\cdot 4^k+17=3\cdot 4^k + 4^k+17

3\cdot 4^k is divisible by 3 and

4^k+17 is divisible by 3, by induction hypothesis

So, the sum is divisible by 3.

Step 3 - Conclusion

We just prove that 4^n+17 is divisible by 3 for all positive integers n.

Thanks

4 0
3 years ago
Rewrite the radical expression as an expression with rational exponents.
natita [175]

Answer:

x^(7/5)

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x^(7/5)

4 0
2 years ago
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Find the minimum or maximum value of the function g(x)=−4x^2+16x−15
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Extra Letters.

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2 years ago
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