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k0ka [10]
3 years ago
11

The equation C = 20n + 35 represents the relationship between the cost of school volleyball uniforms,

Mathematics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

<u>Given the equation </u>

  • C = 20n + 35

<u>We need:</u>

  • n ≥ 15 and C ≤ 600

<u>following</u>

  • 20n + 35 ≤ 600
  • 20n ≤ 565
  • n ≤ 565/20
  • n ≤ 28 (rounded to the whole number)

<u>Considering that n ≥ 15, we have:</u>

  • 15 ≤ n ≤ 28

<u>And the cost is going to be:</u>

  • 20*15 + 35 ≤ C ≤ 20*28 + 35
  • 335 ≤ C ≤ 595
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Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

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Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

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if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

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but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

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4 years ago
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