Answer:
prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S
Explanation:
Given that S = { p q |p, q are prime numbers greater than 0}
E = {0, −2, 2, −4, 4, −6, 6, · · · }
To prove by constructing a bijection from S to E
detailed solution attached below
After the bijection :
<em>prove that | S | = | E |</em> : every element of S there is an Image on E , while not every element on E has an image on S
∴ we can say sets E and S are infinite sets
Answer:
OSHA Hazard Communication Standard is responsible.
Explanation:
The air flow necessary to remain at the lower explosive level is 4515. 04cfm
<h3>How to solve for the rate of air flow</h3>
First we have to find the rate of emission. This is solved as
2pints/1.5 x 1min
= 2/1.5x60
We have the following details
SG = 0.71
LEL = 1.9%
B = 10% = 0.1 a constant
The molecular weight is given as 74.12
Then we would have Q as
403*100*0.2222 / 74.12 * 0.71 * 0.1
= Q = 4515. 04
Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm
Read more on the rate of air flow on brainly.com/question/13289839
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