Answer:
C.Supersaturated
Explanation:
There are three types of solution:
<u>SATURATED SOLUTION</u>:
It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.
<u>UNSATURATED SOLUTION</u>:
It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.
<u>SUPERSATURATED SOLUTION:</u>
It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.
Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:
<u>C.Supersaturated</u>
Answer:
option D
Explanation:
Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.
Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.
Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.
Hence, the correct answer is option D
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) =
Answer:
present
Explanation:
read doesn't change but write is in present tense
