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BaLLatris [955]
3 years ago
7

I need help with this.

Mathematics
1 answer:
likoan [24]3 years ago
5 0

a quick clarification and then some.

Profit is what's leftover after the cost is subtracted from the revenue, or namely, if you sell a product for some amount, say 10, and you sold 100 of those, so you made 10*100 or 1000, that's the revenue or the income coming in, however, in making the product you had to cover some expenses, like if it's clothing, well, you have to buy the garment and saw it, have premises and machines to make the clothing and so on, all that's expenses, and since it's out of pocket money, is Cost, if you subtract that Cost from the 1000 in Revenue, what's leftover, that surplus is Profit.

this company sells "p" units, each at $855, so their Revenue will just be 855p, expenses or Cost is a fixed amount of $6780, so the profit comes from their difference.

on a given month, they sold 250, namely p = 250, so their Revenue is just 250*855 = 213750 bucks, however, we need to remove the Costs of 6780, 213750 - 6780 = 206970, that's our Profit.

now, the CEO takes a bite of 15% of that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{15\% of 206970}}{\left( \cfrac{15}{100} \right)206970}\implies 31045.5

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Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

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To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
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aleksandr82 [10.1K]

Answer:

3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are  (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).

Step-by-step explanation:

 Let the First positive number be <em>x</em> and second positive number be <em>y.</em>

Triple of first number = 3<em>x</em>

double of second number = 2<em>y</em>

According to question,

3<em>x </em>+ 2<em>y</em> = 34

Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34

So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34

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3<em>x </em>= 34 - 2<em>y</em>

Let x = 10 and by substituting its value in above expression,

3 \times 10 = 34 - 2y

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2y = 34 - 30

2y = 4

y = 2

Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)

In the same way put x = 4 then,

3<em>x </em>= 34 - 2<em>y</em>

3 \times 4 = 34 - 2y

2y = 34 - 12

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y = 11

Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)

Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and  (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.  

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How to rewrite the sum of 96 80 using distributive property?
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