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levacccp [35]
3 years ago
13

Helpppppp pleaseeeeee

Mathematics
1 answer:
Natalija [7]3 years ago
3 0

Select the second and fourth options.

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Use the equation e^x+e^-x/e^x-e^-x=t tosolve for x
lana66690 [7]

Answer:

Solve the rational equation by combining expressions and isolating the variable  

<h2>x = ln ( t − √ t ^2 + 4  )/2</h2><h2>x = ln ( t +√ t ^2 + 4  )/2</h2>

4 0
3 years ago
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I need major help pretty pls this so different
crimeas [40]
First put a bubble around 5x-7. Now you plug that into the problem. -3x-2
(5x-7)=12. now times the -2 by 5x which is -10x. and then times the -2 by -7 which is 14. so now you have -3x-10x+14=-12. add the -3x and -10x. -13x. Add the 14 over. -13x=2. Now dived by -13, which gives us x=-0.15

Now you take that a plug it into the -3x-2y=-12. So that it looks like this -3(-0.15)-2y=-12

Now times the -3 by -0.15 to give us 20 -2y=-12.   Now 'kill' the 20, so that it looks like this -2y=-32. Now you divide by -2 and get y=16. 

Now your ordered pair is (-0.15,16)
7 0
3 years ago
Obtén lo que se te pide, dadas las dos matrices siguientes.
Ivan
I think its b because it makes sense
6 0
3 years ago
Solve the formula for one of its variables using addition subtraction and/ or division
Tpy6a [65]

THE ANSWER IS

pi r ^2 / V

or pi r squared over V

5 0
3 years ago
How much of a radioactive kind of ruthenium will be left after 8 hours if you start with 3,888
ziro4ka [17]

Answer:

Step-by-step explanation:

As an example, iodine-131 is a radioisotope with a half-life of 8 days. It decays by beta particle emission into xenon-131. After eight days have passed, half of the atoms of any sample of iodine-131 will have decayed, and the sample will now be 50% iodine-131 and 50% xenon-131.

50 grams to 25 grams is one half-life. 25 grams to 12.5 grams is another half-life. So, for 50 grams to decay to 12.5 grams, two half-lives, which would take 36 days total, would need to pass. This means each half-life for element X is 18 days.

7 0
3 years ago
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