Factor Completely: 4x²+8x

Mathematics

2 answers:

Sidana [21]4 years ago

8 0

The answer is 4x(x+2).

Nitella [24]4 years ago

6 0

I got

4x^2+8x=

x^2=x*x=

4*x*x+8x=

4xx+4*2x=

4x(x+2)

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Sara had 94 dollars to spend on 8 books. After buying them she had 14 dollars. How much did each book cost

cestrela7 [59]

Answer: $10

Step-by-step explanation:

Let's analize the information given: We know that she had $94 to spend on 8 books and after buying these books she had $14. So, we need to calculate the amount of money she spent. Subtract $94 and $14:

$Total\ spent=\$94-\$14\\Total\ spent=\$80$

Then, to calculate the cost of each book, we need to divide the "Total spent" by 8:

$cost\ of\ each\ book=\frac{Total\ spent}{8}\\\\cost\ of\ each\ book=\frac{\$80}{8}\\\\cost\ of\ each\ book=\$10$

3 0

3 years ago

Custom Office makes a line of executive desks. It is estimated that the total cost for making x units of their Senior Executive

Ivan

Answer:

(a) The average cost function is $\bar{C}(x)=95+\frac{230000}{x}$

(b) The marginal average cost function is $\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

Step-by-step explanation:

(a) Suppose $C(x)$ is a total cost function. Then the average cost function, denoted by $\bar{C}(x)$ , is

$\frac{C(x)}{x}$

We know that the total cost for making x units of their Senior Executive model is given by the function

$C(x) = 95x + 230000$

The average cost function is

$\bar{C}(x)=\frac{C(x)}{x}=\frac{95x + 230000}{x} \\\bar{C}(x)=95+\frac{230000}{x}$

(b) The derivative $\bar{C}'(x)$ of the average cost function, called the marginal average cost function, measures the rate of change of the average cost function with respect to the number of units produced.

The marginal average cost function is

$\bar{C}'(x)=\frac{d}{dx}\left(95+\frac{230000}{x}\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g\\\\\frac{d}{dx}\left(95\right)+\frac{d}{dx}\left(\frac{230000}{x}\right)\\\\\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

$\lim_{x \to \infty} (\bar{C}(x))=\lim_{x \to \infty} (95+\frac{230000}{x})\\\\\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)\\\\=\lim _{x\to \infty \:}\left(95\right)+\lim _{x\to \infty \:}\left(\frac{230000}{x}\right)\\\\\lim _{x\to a}c=c\\\lim _{x\to \infty \:}\left(95\right)=95\\\\\mathrm{Apply\:Infinity\:Property:}\:\lim _{x\to \infty }\left(\frac{c}{x^a}\right)=0\\\lim_{x \to \infty} (\frac{230000}{x} )=0$