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Scilla [17]
3 years ago
14

Scenario: The Cannon Instructions: View the video found on page 1 of this Journal activity. Using the information provided in th

e video, answer the questions below. Show your work for all calculations. The Cannon: Ernest's friend Nik is about to be shot out of a cannon. The path he will travel follows a parabolic arch that can be described by this polynomial. f(x) = –0.05(x2 – 26x – 120) He is supposed to land on a safety net 30 feet away. Does the function give you enough information to tell you where he will land? If so, how far from the cannon will he land?
Mathematics
1 answer:
Artyom0805 [142]3 years ago
3 0

Answer:

Part A

Yes

Part B

He will land at a point 30 meters from Cannon

Step-by-step explanation:

Part A

The given function that represent the path of the Cannon, which is the path of a parabolic arc, is presented as follows;

f(x) = -0.05·(x² - 26·x - 120)

We note that the height, 'h', of the path of a parabola at a horizontal distance, 'x' from the origin is equal to f(x)

Therefore, given that at the ground level, at the start and end of the flight, the f(x) = 0, we can write;

-0.05·(x² - 26·x - 120) = 0

∴ x² - 26·x - 120 = 0

x = (26 ± √((-26)² - 4 × 1 × (-120)))/(2 × 1)

Which gives;

x = 30 or x = -4

Therefore, he lands at a point 30 meters from the starting point

Therefore, the function gives enough information to tell where he will land

Part B

He lands at the point, x = 30 from the point the cannon was fired.

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Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f
GalinKa [24]

Answer:

\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}

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We identify the coefficients for the like terms, it comes

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\\f(x)=(x^2+16)(x^2-2x-15)

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f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}

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f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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Step-by-step explanation:

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