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3 years ago

If increasing 15 by n% is equivalent to decreasing 75 by n%, then what is 6% of n?

1 answer:

4 0

15(1 + n/100) = 75(1 - n/100)
15 + 0.15n = 75 - 0.75n
0.15n + 0.75n = 75 - 15
0.9n = 60
n = 60/0.9 = 66 2/3%

6% of n = 0.06 x 66 2/3 = 4

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(3/5x + 4) + (1/5x -1)

kiruha [24]

Answer:

4/5x+3

Step-by-step explanation:

remove the useless parenthesis, then calculate the x sum then the normal numbers sum

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.

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

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

6 0

3 years ago

Find GCF of -50m ^4 n ^7 and 40m ^2 n ^10

trapecia [35]

-50m^4 n^7 = 10m^2n^7(-5m^2)<span>
and
40m^2 n^10 = 10</span>m^2 n^7(4n^3)

Answer: GCF = 10m^2 n^7

5 0

3 years ago

An isosceles trapazoid has base angles of 45 degrees and bases of lengths 9 and 15. The area of the trapazoid is what

lakkis [162]

Answer:

area of trapazoid= 1/2 × b× h

so = 1/2×9×15

=1/2×135

=67.5cm

6 0

2 years ago

In ΔABC, the measure of ∠C=90°, the measure of ∠B=64°, and BC= 6.6 Find the length of CA the nearest tenth of a foot

Marrrta [24]

Answer:

CA = 13.5

Step-by-step explanation:

tan 64 = CA/6.6

2.05 = CA/6.6

CA = 13.5

4 0

3 years ago