I NEED HELP ASAP PLEASE. I CANT THINK TOO MUCH RN SO YEAH PLEASE HELP

Mathematics

1 answer:

BaLLatris [955]3 years ago

4 0

There are 60 min per hour.
There are 24 hours in one day, so there are 60 times 24= 1,440 minutes in a day.
Person A: for 45 minutes per day had a heart rate of 150, and for 1,395 minutes per day had a heart rate of 60.
This means that her heart rate totaled (45x150) + (1,395x60) = 90,450 beats per day.
Person B: for 1,440 minutes per day had a heart rate of 72 beats per minute.
This means that her heart rate totaled (1,440x72) = 103,680 beats per day.

So for 365 days: 365(103,680-90,450)= 4,828,950-this is the number of beats MORE per year of Person B than Person A.

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elena-s [515]

Are you okay? Why the spam?

7 0

3 years ago

Read 2 more answers

Pls help this is pretty urgent

laiz [17]

Answer:

(a) See below.

(b) x = 0 or x = 1

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{\ln |x-1|}{x}$

Part (a)

Substitute x = 2 into the given rational function:

$\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}$

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

Part (b)

Given interval: [-2, 2]

Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, the numerator is undefined when:

|x - 1| = 0 ⇒ x = 1.

A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.

So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].

Part (c)

x = 1 is a vertical asymptote. As the function exists on both sides of this vertical asymptote, it is an infinite discontinuity. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.

A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.

$\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}$

Therefore, there is a hole at x = 0.

The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it. Therefore, x = 0 is a removable discontinuity.