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3 years ago

How can you represent real world situations using positive nubers, negative nubers, and zero?

1 answer:

Natasha2012 [34]3 years ago

6 0

Answer:Maybe ffor the positive numbers you could use like 1,2,3,4,5,6,7,8, and 9. For the negitive number could be like -1,-2,-3,-4,-5. And zero could repersent the middle. Get it or no?

Step-by-step explanation:

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3 years ago

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Please help!!!! <br><br> If f(1) = 2 and f(n+1) = f(n)^2 – 3 then find the value of f(3).

Black_prince [1.1K]

Answer:

The value of f(3) is -2.

Step-by-step explanation:

This is a recursive function. So

$f(1) = 2$

Now, we find f(2) in function of f(1). So

$f(1+1) = f(1)^2 - 3$

$f(2) = f(1)^2 - 3 = 2^2 - 3 = 1$

Now, with f(2), we can find the value of f(3).

$f(2+1) = f(2)^2 - 3$

$f(3) = f(2)^2 - 3 = 1^2 - 3 = -2$

The value of f(3) is -2.

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3 years ago

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

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3 years ago

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Ilia_Sergeevich [38]

The statement is true.

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