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hoa [83]
2 years ago
11

What is the leading coefficient of a third degree function that has an output of 221 when x=2, and has zeros of −15, 3i, and −3i

?
Mathematics
1 answer:
Vinil7 [7]2 years ago
3 0

Answer:

  1

Step-by-step explanation:

The function with the given zeros will factor as ...

  f(x) = a(x +15)(x^2 +9) . . . . with leading  coefficient 'a'

You have ...

  f(2) = 221 = a(2+15)(2^2+9) = a(17)(13) = 221a

Then a = 221/221 = 1

The leading coefficient is 1.

_____

<em>Additional comment</em>

As you know, a function with zero x=p has a factor of (x -p). The given zeros mean the function has factors (x -(-15)), (x -3i). and (x -(-3i)). The product of the last two factors is the difference of squares: (x^2 -(3i)^2) = (x^2 -(-9)) = (x^2 +9). This is how we arrived at the factorization shown above.

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<h3><u>Solution:</u></h3>

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Then from given statement,

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<em><u>Substituting given values we get,</u></em>

6 \times 10^{22} = x \times 3000\\\\x = \frac{6 \times 10^{22}}{3000}\\\\x = \frac{6 \times 10^{22}}{3 \times 10^3}\\\\x = 2 \times 10^{22-3}\\\\x = 2 \times 10^{19}

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Answer:

Step-by-step explanation:

here you go mate

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