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zlopas [31]
3 years ago
12

A company that makes rubber ducks has a start up cost of $ 7. It costs the company $ 2.91 to make each rubber duck. The company

charges $ 3.74 for each rubber duck. Let denote the number of rubber ducks produced. Determine the cost function for this company. The cost function is the amount it costs to manufacture the rubber ducks. C(x)=
Mathematics
1 answer:
leonid [27]3 years ago
8 0
A start up cost for a company is $7.
It costs the company $2.91 to make each rubber duck.
The amount of ducks is q.
We have to determine the cost function for this company ( and $3.74 is connected with the revenue, not with the costs ).
Answer:
C ( x ) = 7 + 2.91 q.
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A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)
Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

A)

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by <em>x</em> ²,

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}

Then each remaining rational term converges to 0 as <em>x</em> gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than <em>x</em> - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}

More precisely,

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0

D) Looks like this limit should read

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}

which is just another case of (A) and (B); the limit would be

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1

That is,

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1

However, in case you meant something else, such as

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}

then the limit would be different:

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0

since the degree of the denominator is larger.

One important detail glossed over here is that

\sqrt{t^2} = |t|

for all real <em>t</em>. But since <em>t</em> is approaching *positive* infinity, we have <em>t</em> > 0, for which |<em>t</em> | = <em>t</em>.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}

Again,

\sqrt{x^8} = |x^4|

but <em>x</em> ⁴ is non-negative for real <em>x</em>.

F) Also somewhat ambiguous:

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3

or

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large <em>x</em>, <em>x</em> ⁴ is much larger than <em>x</em> ², so effectively

\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}

6 0
2 years ago
I need help with this
Lisa [10]

Answer:

1

Step-by-step explanation:

15,580.84 rounded to the nearest 10 thousand is 20,000

8 0
3 years ago
If your teacher gave you instruction for constructing a regular pentagon inscribed in a circle, you would expect the pentagon to
zaharov [31]

Touching the perimeter

Hope this helps :)


3 0
3 years ago
From her purchased bags, Rory counted 110 red candies out of 550 total candies. Using a 90% confidence interval for the populati
Svetlanka [38]

Answer:

The Confidence Interval = (0.172, 0.228)

Where:

The lower limit = 0.172

The upper limit = 0.228

Step-by-step explanation:

The formula to be applied or used to solve this question is :

Confidence Interval formula for proportion.

The formula is given as :

p ± z × √[p(1 - p)/n]

n = Total number of red candies = 550 red candles

p = proportion = Number of red candies counted/ Total number of red candies

= 110/550 = 1/5 = 0.2

z = z score for the given confidence interval.

We are given a confidence interval of 90%. Therefore, the z score = 1.6449

Confidence Interval = p ± z × √[p(1 - p)/n]

Confidence Interval = 0.2 ± 1.6449 × √[0.2(1 - 0.2)/550]

= 0.2 ± 1.6449 √0.2 × 0.8/550

= 0.2 ± 1.6449 × 0.0170560573

= 0.2 ± 0.0280555087

Hence, the Confidence Interval = 0.2 ± 0.0280555087

0.2 - 0.0280555087 = 0.1719444913

Approximately = 0.172

0.2 + 0.0280555087 = 0.2280555087

Approximately = 0.228

Therefore, the Confidence Interval = (0.172, 0.228)

Where:

The lower limit = 0.172

The upper limit = 0.228

4 0
3 years ago
Funny questions
melamori03 [73]

Answer:

Hey lol.

I would  visit the future

home in the mountains

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Step-by-step explanation:

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