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emmainna [20.7K]
2 years ago
6

I need help solving this

Mathematics
1 answer:
aleksandrvk [35]2 years ago
8 0

Answer:

45.45%

Step-by-step explanation:

freshman and sophomores over total students

(110+65)/(110+65+95+50)

175/385

45.45%

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Q is a degree 5 polynomial that passes through the origin, has zeros i and 7 - i, and q(-1)= -520. find the equation for q​
Lunna [17]

Using the factor theorem, the equation for q is:

q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)

The <em>Factor Theorem</em> states that a polynomial function with roots x_1, x_2, ..., x_n is given by:

f(x) = a(x - x_1)(x - x_2)...(x - x_n)

In which a is the leading coefficient.

In this problem:

  • Passes through the origin, so x = 0 is a zero, that is x_1 = 0.
  • x = i is a zero, and thus, it's conjugate also is, that is, x = -i, hence x_2 = i, x_3 = -i
  • x = 7 - i is a zero, and thus, it's conjugate also is, that is, x = 7 + i, hence x_4 = 7 - i, x_3 = 7 + i

Then, the equation is:

q(x) = a(x - 0)(x - i)(x + i)(x - 7 + i)(x - 7 - i)

q(x) = ax(x^2 - i^2)((x-7)^2 - i^2)

Considering that i^2 = -1

q(x) = ax(x^2 + 1)(x^2 - 14x + 50)

q(x) = ax(x^4 - 14x^3 + 51x^2 - 14x + 50)

q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)

q(-1) = -520 means that when x = -1, q = -520, and this is used to find a.

q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)

-520 = a(-1 - 14 - 51 - 14 - 50)

130a = 520

a = \frac{520}{130}

a = 4

Hence, the equation is:

q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)

A similar problem is given at brainly.com/question/24380382

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