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Valentin [98]
4 years ago
7

I need help!!!

Mathematics
1 answer:
Kruka [31]4 years ago
4 0

Answer: x = -3/4 can not be a rational zero of the polynomial.

Step-by-step explanation:

We have the polynomial:

6x^5 + ax^3 -bx -12 = 0.

The theorem says that:

If P(x) is a polynomial with integer coefficients, and p/q is a zero of P(x) then p is a factor of the constant term (in this case the constant term is -12) and q is a factor of the leading coefficient (in this case the leading coefficient is 6.).

The factors of -12  (different than itself) are (independent of the sign).

1, 2, 3, 4 and 6.

So p can be: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6.

The factors of 6 are:

1, 2 and 3, so q can be 1, -1, 2, -2, 3, -3.

Then the option that can not be a zero of the polynomial is

x = -3/4

because the number in the denominator must be a factor of the leading coefficient, and 4 is not a factor of six.

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3 years ago
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Step-by-step explanation:

A few formulas an definitions which will help us:

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A=\pi r^2\\=\pi\left(\frac{d}{2}\right)^2\\=\pi\left(\frac{d^2}{4}\right)\\=\pi\left(\frac{(c/\pi)^2}{4}\right)\\=\pi\left(\frac{c^2/\pi^2}{4}\right)\\=\pi\left(\frac{c^2}{4\pi^2}\right)\\\\=\frac{\pi c^2}{4\pi^2}\\ =\frac{c^2}{4\pi}

We can now substitute c for our circumference, 65, to get our answer in terms of π:

A=\dfrac{65^2}{4\pi}=\dfrac{4225}{4\pi}=1056.25\pi

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