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2 years ago

DO IT ASAP! Thank you very much!

Mathematics

1 answer:

frozen [14]2 years ago

3 0

Answer:

a = 5

Step-by-step explanation:

According to the factor theorem, if x + 2 is a factor, then by dividing the polynomial by the binomial, we are meant not to have a remainder

In this case, the remainder would be zero

So, if we set the binomial equals zero and substitute the x-value into the polynomial, we are supposed to have 0

So we have this as;

x + 2 = 0

x = -2

-2^3 -2(-2)^2 -2(a) + 6 = 0

-8-8-2a + 6 = 0

-16 + 2a + 6 = 0

2a -10 = 0

2a = 10

a = 10/2

a = 5

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Find the distance between the points (-1,-9) and (1,-4)

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The distance between the two points is 5.4

Explanation:

The points are (-1,-9) and (1,-4)

The distance between the two points can be determined using the formula

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Substituting the values in the formula, we get,

$d=\sqrt{(1+1)^{2}+(-4+9)^{2}}$

Adding the values,

$d=\sqrt{2^{2}+5^{2}}$

Squaring the terms, we get,

$d=\sqrt{4+25}$

Simplifying, we have,

$d=\sqrt{29}=5.4$

Thus, the distance between the two points is 5.4

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Read 2 more answers

Evaluate the infinite sum:

satela [25.4K]

Consider the k-th partial sum,

$S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}$

More compactly,

$\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}$

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as k goes to infinity. We have

$\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}$

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |x| < 1, we have

$\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n$

Differentiating both sides gives

$\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}$

also valid for |x| < 1. Take x = 1/π and you get the sum you want to compute.

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2 years ago