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Alona [7]
4 years ago
5

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

Mathematics
1 answer:
pantera1 [17]4 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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Manuel hizo un viaje en el coche, en el cual consumió 20 litros de gasolina. El trayecto lo hizo en dos partes: en la primera, c
Alona [7]

Responder:

24 litros; 16 litros; 4 litros

Explicación paso a paso:

Dado que:

Gasolina consumida = 20 litros

Sea la cantidad de gasolina en el tanque = x

Primera parte del viaje = 2/3 de x

Segunda parte del viaje = 1/2 de (x - 2x / 3)

Cantidad de gasolina en el tanque:

2x / 3 + 1/2 (x - 2x / 3) = 20

Solución para x

2x / 3 + x / 2 - x / 3 = 20

(4x + 3x - 2x) / 6 = 20

5 veces / 6 = 20

5 veces = 20 * 6

5 veces = 120

x = 120/5

x = 24

Cantidad de gasolina en el tanque = 24 litros

Litros consumidos en cada etapa:

Primera parte = 2/3 de 24 = 48/3 = 16 litros

2a parte = 0.5 de (24 - 16) = 0.5 * 8 = 4 litros

3 0
3 years ago
Darin has 5 coins that the total 62%. What are the coins?
Snowcat [4.5K]
Hi there!

There are several possible options.

A main one would be : 10 DIMES and 2 PENNIES

Other ones could be :

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1 dime
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62 pennies

12 nickels
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5 0
3 years ago
Help please bed asap thank you!!!
Mkey [24]

Answer:

hr

Step-by-step explanation:

8 0
2 years ago
PLEASE HELP!!!!!!!!!!
Oksanka [162]

Answer:

B

Step-by-step explanation:

Data set D does not contain the value 128, which is the median value.

Data set C does not contain the outlier value 91.

Data set A contains value 168, which does not show up on the plot.

The only remaining choice is B.

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In order, the data values of set B are ...

... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152

The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)

6 0
4 years ago
Read 2 more answers
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Mkey [24]
The answer to c. Is 8t+12g
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