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Alex_Xolod [135]

3 years ago

7

Explain how you can use the terms from the binomial expansion to approximate 0.985.

1 answer:

stiv31 [10]3 years ago

5 0

There are 5 + 1 = 6 terms in the binomial expansion of (1−0.02)5, and since the 4th term is approximately 0, the 5th and 6th terms are also approximately 0. So, approximate the value of 0.985 by adding the first three terms: 1 + (-0.1) + 0.004 = 0.904.

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Black_prince [1.1K]

Answer:

The solutions of the quadratic equation are $x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}$

Step-by-step explanation:

This is a second order polynomial, and we can find it's roots by the Bhaskara formula.

Explanation of the bhaskara formula:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

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$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

For this problem, we have to find $x_{1} \text{and} x_{2}$ .

The polynomial is $3x^{2} - 5x +1$ , so a = 3, b = -5, c = 1.

Solution

$\bigtriangleup = b^{2} - 4ac = (-5)^{2} - 4*3*1 = 25 - 12 = 13$

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) + \sqrt{13}}{2*3} = \frac{5 + \sqrt{13}}{6}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) - \sqrt{13}}{2*3} = \frac{5 - \sqrt{13}}{6}$

The solutions of the quadratic equation are $x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}$

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