The shape of Earth's lithosphere is continually changing. Bedrock that is closer to oceanic ridges is younger in age than bedrock that is farther away. What can best be concluded from this information? It can cause pollution of water in lakes and rivers
Answer:
10⁴¹ s quark top lives have been in the history of the universe.
Explanation:
You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:
Yo know that the "Age of the universe" is 100,000,000,000,000,000 which can also be expressed as 10¹⁷ s
.
You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.
Then 
Recalling that the result of dividing two powers of the same base is another power with the same base where the exponent is the subtraction of the initial exponents, it is possible to calculate this division as follows:
<u><em>t=10⁴¹ s</em></u>
So <u><em>10⁴¹ s quark top lives have been in the history of the universe.</em></u>
Answer: the answer is A, C, E
Explanation:
Time management is apart of your study schedule so you will need to determine when to study, same goes with how to use ur study time, lastly it helps create ur study schedule to help find out what time you are free to study. :)
The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
