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jekas [21]
3 years ago
7

Xin Used 20 yards of fencing to build the walls of a square a chicken coop. Used 20 yards of fencing to build the walls of a squ

are chicken coop Which equation and solution represent X , The length , in yards , Of each wall oF the square coop
Mathematics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

Equation: 20=4x or x=\frac{20}{4}

Solution: x=5

Step-by-step explanation:

The formula that is used to calculate the perimeter of  a square is:

P=4s

Where "s" is the side lenght the square.

Knowing that Xin used  20 yards of fencing to build the walls of a square chicken coop (in which the lenght in yards of each wall is represented with "x"), you can identify that:

P=20\\\\s=x

Then, you can susbtitute values into the formula P=4s:

20=4x

Finally, you must solve for "x" in order to find its value.

This is:

\frac{20}{4}=x\\\\x=5

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3 years ago
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Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

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)}\\

And then, we use the chain rule to differentiate:

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

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:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

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)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

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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