Prove that if a is a natural number, then there exist two unequal natural numbers k and l for which ak − al is divisible by 10.

Sixty percent of Company A’s employees are considered top performers. Fifty five percent of Company A’s employees are in a rigor

pickupchik [31]

Not exactly. there are still 40 percent of the employees that are not top performers. for all you know those 40 percent could make up the majority of the employees in the training program<span />

5 0

2 years ago

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What are the factors of the polynomial function? Use the rational root theorem to determine the factors. f(x) = 2x³ + x² - 8x -

fredd [130]

Answer:

( x + 2 ) ( x - 2 ) ( 2x + 1)

Step-by-step explanation:

coefficient of x³ is 2 and denote it with q and denote -4 as p

the set of possible rational roots through rational theorem will be within ± (p/q)

now factors of -4 are ±(1,2,4) and factors of 2 are ± (1,2) the possible rational roots are ± ( 1/1, 1/2, 2/1, 2/2, 4/1, 4/2) which reduces to ± (1, 1/2, 2, 4)

substitute each of the value into the equation

f(x) = 2x³ + x² - 8x - 4 to the root ( that gives f(x) = 0)

the equation can be made easy writing it in reduced form

2x³ + x² - 8x - 4

(2x³ + x²) - (8x + 4)

x² (2x + 1) - 4 (2x + 1)

(x² - 4) (2x + 1)

( x + 2 ) ( x - 2) ( 2x + 1) are the factors which correspond to 2, -2, -1/2 roots

4 0

2 years ago

Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$ >0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that $|a_n-A|n_0$ .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we denote the first series as $b_n=1$ , we will have that $\lim_{k \to \infty} b_k=1$ .

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$ .

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$ .

Since those limits of sub-series are different, we can never find such $n_0\\$ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose $\epsilon =1$ , all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$ .