Round 49.87654321 to the nearest tenth

The "int()" operator in your calculator is the "greatest integer function." It returns the greatest integer less than or equal t

Leona [35]

The function that correctly returns grades from 60 percent to a 100 percent is f(x) = [[(x - 50)/10]]

<h3>How to determine this function</h3>

The question tells us that the students do not fail and they do not get a 100 percent score either.

Hence there is no need for worry if the grade is at 5.0 or less than 1.

For example we have

x = 82.976

f(82.976) =[[(82.976-50)/10]

This would then be 3.29

This is then = 3.0

This function is not an odd function and it is not an even function. This is because the domain does not have negative values.

Read more on the domain of a function here:

brainly.com/question/1369616

#SPJ1

5 0

2 years ago

Trey had a cylindrical can with a radius of 4 inches and a height of 12 inches that was completely full of beet juice. He poured

Slav-nsk [51]

Answer:

$8\ glasses\ of\ juice$

Step-by-step explanation:

step 1

Find the volume of the cylindrical can

The volume of the cylinder is equal to

$V=\pi r^{2}h$

we have

$r=4\ in$

$h=12\ in$

substitute the values

$V=\pi (4)^{2}(12)=192\pi\ in^{3}$

step 2

Find the volume of the cylindrical glasses

The volume of the cylinder is equal to

$V=\pi r^{2}h$

we have

$r=2\ in$

$h=8\ in$

substitute the values

$V=\pi (2)^{2}(8)=32\pi\ in^{3}$

<em>Find the volume of 3/4 of the glass</em>

$32\pi (\frac{3}{4})=24\pi\ in^{3}$

step 3

Divide the total volume of cylindrical can by the 3/4 volume of the glass

$192\pi\ in^{3}/24\pi\ in^{3}=8\ glasses\ of\ juice$

5 0

3 years ago

(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also c

valina [46]

Answer:

a) See the proof below.

b) $\sum \frac{(-x)^n}{n}$

Step-by-step explanation:

Part a

For this case we assume that we have the following series $\sum a)n x^n$ and this series has a finite radius of convergence $R$ and we assume that $a_n \geq 0$ for all n, this information is given by the problem.

We assume that the series converges at the point $x= R$ since w eknwo that converges, and since converges we can conclude that:

$\sum a)n R^n < \infty$

For this case we need to show that converges also for $x=-R$

So we need to proof that $\sum a_n (-R)^n < \infty$

We can do some algebra and we can rewrite the following expression like this:

$\sum a_n (-R)^n = \sum (-1)^n a)n R^n$ and we see that the last series is alternating.

Since we know that $\sum a_n x^n$ converges then the sequence { $a_n R^n$ } must be positive and we need to have $lim_{n\to \infty} a^n R^n = 0$

And then by the alternating series test we can conclude that $\sum a_n (-R)^n$ also converges. And then we conclude that the power series $a_n x^n$ converges for $x=-R$ ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this $\frac{(-x)^n}{n}$

Because the series $\sum \frac{(-x)^n}{n}$ converges to $-ln(1+x)$ when $|x|$ using the root test.

But by the properties of the natural log the series diverges at $x=-1$ because $\sum \frac{1}{n} =\infty$ and for $x=1$ we know that converges since $\sum \frac{-1}{n}$ is an alternating series that converges because the expression tends to 0.