The equation y = 0.75x represents the cost of watermelon in dollars (y) and the number of pounds of watermelon(x). What is the c
1 answer:
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Answer:
49.7%
Step-by-step explanation:
A cdircle is located within a square.
<u>Area of the circle</u>
Area = , where r = 4 units.
Area Circle = 50.3 units^2
<u>Area of the square</u>
Area = l*w or l^2 for a square, since l = w
Area = (10 units)^2
Area = 100 units^2
<u>Area in the square but outside the circle</u>
This is the difference [Square minus Circle Areas]
Square minus Circle Areas = 100 - 50.3 or <u>49.7 units^2</u>
<u>Probability</u>
The probability of picking a point in the square that is not in the circle is the ration of the two areas: <u>[Outside Circle/Square]x100%</u>
<u></u>
<u>(</u>49.7 units^2)/(100 units^2)x100% = 49.7%
<u></u>
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):
The process will help us to understand what is happening, at first we rewrite the function:
And then, we use the chain rule to differentiate:
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)
