The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)

1 answer:

3 0

$\bf ~\hspace{10em}\textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 8x^3-125~~ \begin{cases} 8=2\cdot 2\cdot 2\\ \qquad 2^3\\ 125=5\cdot 5\cdot 5\\ \qquad 5^3 \end{cases}\implies 2^3x^3-5^3\implies (2x)^3-5^3 \\[2em] [2x-5][(2x)^2+(2x)(5)+5^2]\implies (2x-5)(\stackrel{\stackrel{j}{\downarrow }}{4}x^2+\stackrel{\stackrel{k}{\downarrow }}{10}x+25)$

You might be interested in

A zookeeper has 8 frogs. Each frog eats 3/4 cups of food. How much food does the zookeeper need in total?

Kruka [31]

Answer:

can we see the problem please

7 0

3 years ago

The cost to make a Nike shoe is $20 USD and the store wants to make a profit of 75% of the cost price.

Alenkasestr [34]

so 75%of 20 is 15.00

and then we 20.00+15.00= 35.00

plus sales tax of 5%

so I got 36.75$ after taxs

4 0

3 years ago

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,564 miles, with a standard

DerKrebs [107]

Answer:

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$