Can somebody help me ??

1 answer:

5 0

Answer:

i think its 12 i could be wrong so sorry if i am but it might be 12

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Sati [7]

$S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\a_1=6;\ n=56;\ a_{56}=391\\\\subtitute\\\\S_{56}=\dfrac{6+391}{2}\cdot56=\dfrac{397}{2}\cdot56=397\cdot28=11,116$

7 0

3 years ago

laiz [17]

Good evening,

Step-by-step explanation:

Look at the photo below for the Answer.

5 0

3 years ago

Marina CMI [18]

a. There are 0 or 2 real positive roots for the equation and

b. There are 0 or 2 real negative roots for the equation.

<h3>What is the Descartes'rule of sign?</h3>

Descartes' rule of sign states that

The number of real positive zero of a polynomial f(x) is the number of sign changes of the coefficients of f(x) or an even number less than the number of sign changes of the coefficients of f(x)
The number of real negative zero of a polynomial f(x) is the number of sign changes of the coefficients of f(-x) or an even number less than the number of sign changes of the coefficients of f(-x)

<h3>How to find the number of possible positive and negative roots are there for the equation?</h3>

Given the equation 0 = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1, writing it as a polynomial function, we have f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1

<h3>a. The number of positive roots</h3>

So, to find the number of positive roots, we find the number of sign changes of the polynomial f(x).

So, f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1

Since f(x) has coefficients -8, -2, + 8, -4, -1, there are two sign changes from -2 to + 8 and from + 8 to -4.

So, there are 2 or 2 - 2 = 0 real positive roots.

So, there are 0 or 2 real positive roots for the equation.

<h3>b. The number of negative roots</h3>

So, to find the number of negative roots, we find the number of sign changes of the polynomial f(-x).

So, f(-x) = −8(-x)¹⁰ − 2(-x)⁷ + 8(-x)⁴ − 4(-x)² − 1

= −8x¹⁰ + 2x⁷ + 8x⁴ − 4x² − 1