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julia-pushkina [17]
1 year ago
15

Brad has $12 more than John. Together they have $84

Mathematics
2 answers:
ludmilkaskok [199]1 year ago
6 0

Hello,

John : n

Brad : n + 12

n + n + 12 = 84

2n + 12 = 84

2n = 84 - 12

2n = 72

n = 72/2

n = 36

John $36 and Brad $36 + $12 = $48

mario62 [17]1 year ago
6 0

Answer:

jhon =36 brad=48

Step-by-step explanation:

B=J+12

J+B=84

12+J+J=84

2J=72

J=36

B=J+12=36+12=48

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2 years ago
Se van a unir dos tramos de cadena: uno mide 0.75 m de largo y el otro 15/100 de metro . ¿Cual será la longitud total ya que est
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Step-by-step explanation:

It is given that,

The length of section 1 is 0.75 m and the length of section 2 is 15/100 m. The total length of the section is equal to the sum of section 1 and section 2. So, we will add them as follows :

l=0.75\ m+\dfrac{15}{100}\ m\\\\\because \dfrac{15}{100}=0.15\\\\l=0.75\ m +0.15\ m\\\\l=0.9\ m

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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

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Step-by-step explanation:

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