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3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Mathematics

1 answer:

pantera1 [17]3 years ago

7 0

Answer:

It diverges.

Step-by-step explanation:

We are given the integral: $\int\limits^{-1}_{-\infty} \ln |x| dx$

$\int\limits^{-1}_{-\infty} \ln |x| dx=\int\limits^{-1}_{-\infty} \ln (-x) dx=\\\\= \lim_{t \to \infty} \int\limits^{-1}_{-t} \ln (-x) dx= \lim_{t \to \infty}( x(\ln \left(-x\right)-1))|^{-1}_{-t}=1-\infty=-\infty$

So it is divergent.

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2 years ago

An expression is shown below:

PolarNik [594]

Notation
I imagine that the expression you are asked to work with is:
$3 x^{3}y+15xy-9 x^{2} y-45y$

When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.

PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.

This gives us: $3 x^{3}y+15xy-9 x^{2} y-45y=3y( x^{3}+5x-3 x^{2} -15)$

To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.

PART B
Start with the answer in part A. Namely, $3y( x^{3}+5x-3 x^{2} -15)$ . For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets. $[( x^{3}+5x)]-[3 x^{2} -15]$

Let's next factor what is in each bracket:
$[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]$

Notice that both brackets have the same expression in them so now we factor that out: $[x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)$

Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: $3 x^{3}y+15xy-9 x^{2} y-45y =(3y) (x-3)( x^{2} +5)$

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3 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are arranged to form four two-digit numbers. What is the largest amount of primes that co

artcher [175]

<h3>Answer: 3 primes</h3>

==========================================================

Reason:

The only even prime number is 2. The other primes are odd.

The list of the first few primes are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43

A prime is any number that has factors of 1 and itself, and nothing else.

For instance, 13 is prime because 1 and 13 are the only factors.

-----------

As you can see, if we want a two digit prime number, then the units digit must be odd. Otherwise, 2 is a factor making it not prime (aka composite).

So far we see that the units digit is 1, 3, 5, 7, or 9. But wait, if 5 is the units digit then 5 is a factor. Eg: 5 is a factor of 35 since 5*7 = 35. Refer to the divisibility by 5 rule.

So we reduce the list of choices to 1, 3, 7, 9 for the units digit.

Unfortunately 9 is not in the list of original numbers given, so we can't use it. We really have 1, 3, or 7 as our choices to form the units digit of the two-digit number.

-----------

Let's try to build some primes.

Pick the smallest odd number 1 as the units digit. Then pick 2 as the tens digit. The number 21 is composite because 21 = 7*3, so we rule it out.

On the other hand, 31 is prime since only 1 and 31 are factors.

The problem is that now "3" is tied up and cannot be used for another prime. The good news is that 41 is prime and that's what I'll go with.

Cross 1 and 4 off the list.

The next odd number is 3 which is our units digit. The value 23 is prime.

So far we have 41 and 23 as our two primes.

Like mentioned earlier, we cannot use 5 as the units digit. The number 57 is not prime because 57 = 19*3. So we'll skip over 5.

The number 67 is prime for similar reasoning mentioned earlier.

------------

We have these primes: 41, 23, 67

The next number either 58 or 85 isn't prime

This is one way to show an example of why we're only able to get 3 primes out of this.

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2 years ago