Make a general rule about scale factor, perimeter ratio, and area ratio.

Mathematics

1 answer:

evablogger [386]3 years ago

5 0

Answer: A Scale factor is a number which multiplies or scales a number in an equation.

Perimeter ratio is the ratio of the perimeter to the area.

Area ratio is the ratio of the total amount of usable floor a land has

Step-by-step explanation:

Scale ratio can be established by this simple equation y=CX where C is the scale factor.

Perimeter ratio; perimeter = 2(L+B) where L is length and B is breadth

Area = L×B

Therefore perimeter ratio = 2(L+B)/L×B

Area ratio; L2 ×W2/L1×W1 where L1 and L2 are the lengths and W1 and W2 are the widths.

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4 years ago

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)