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AfilCa [17]
4 years ago
5

Factor f(x)=4x²-15x-25 and then use zero-product property to find the zeros of the quadratic function. Show all work (Will Mark

Brainliest) [Pre Calculus]
Mathematics
1 answer:
Ostrovityanka [42]4 years ago
6 0
<span>f(x)=4x²-15x-25
</span><span>f(x)=(4x + 5)(x - 5)

</span><span>zero-product property 
let </span>f(x)=(4x + 5)(x - 5) = 0
so
(4x + 5)(x - 5) = 0
4x +5 = 0
4x = -5
  x = -5/4
  x = -1.25

x - 5 =0
x = 5

answer
factor: (4x + 5)(x - 5)
x = 5 and x = -1.25

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Answer:

We want to find:

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Here we can use Stirling's approximation, which says that for large values of n, we get:

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Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

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\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

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7 0
3 years ago
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