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lesantik [10]
2 years ago
8

The graph of the function B is shown below. Find B(2), 0.1 01 02

Mathematics
1 answer:
FinnZ [79.3K]2 years ago
8 0

Answer:

i think but not sure it is answer 0.1

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Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-
AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

a)

\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\

So, this integral is convergent.

b)

\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\

So, this integral is divergent.

c)

\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\

So, this integral is divergent.

4 0
3 years ago
Use the quadratic formula to solve for x 8x^2+7x+1=0 plz help
vagabundo [1.1K]

Answer:

X= -7 ± √17 / 16

x= - 0.7798...

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
12-(-19)= <br><br>HELP ME OUT PLZ ​
Vikentia [17]

Answer:

31

Step-by-step explanation:

A negative negative is a positive

12- (-19)

12+19

31

6 0
3 years ago
Write an equation of the line through the given points<br><br>(0, -2), (3, -2)<br>​
KengaRu [80]
Y= my+b
-2=0/3x -2

I’m actually not very sure about this but
8 0
3 years ago
5a + 7 - 8b - 3a +2b - 4
kifflom [539]

━━━━━━━☆☆━━━━━━━

▹ Answer

<em>2a + 3 - 6b</em>

<em />

▹ Step-by-Step Explanation

5a + 7 - 8b - 3a + 2b - 4

<u>Collect like terms</u>

2a + 7 - 8b + 2b - 4

<u>Final Answer</u>

2a + 3 - 6b

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

3 0
3 years ago
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