<u>Approximately 20 million new cases of STIs occur every year in United</u> <u>States.Half of the new cases occur in young people aged between 15–24. </u>
Though younger people are accountable for nearly half of new cases, a recently taken study showed that last year, only around 12% underwent the test for STIs. Center for Disease Control and Prevention (CDC) estimates that each year, undiagnosed STIs causes 24,000 women to become infertile.
Sexually transmitted infections (STIs) are a substantial health challenge which the United States is facing. A very Strong public health awareness and infrastructure is crucial to prevent and control STIs, especially among the younger generation.
Answer:
I think it is the last one.
Explanation:
I am not sure because i am stuck on this one, too.
English: Cardiovascular efficiency depends on a number of factors. One measure is called stroke volume, which is the volume of blood pumped per heartbeat. A fit individual has a larger stroke volume, which means a greater volume of oxygen is delivered to the body per heartbeat.
Spanish: La eficacia cardiovascular depende de una serie de factores. Una medida se denomina volumen sistólico, que es el volumen de sangre bombeada por latidos cardíacos. Un individuo en forma tiene un volumen de movimiento mayor, lo que significa que un mayor volumen de oxígeno es entregado al cuerpo por latidos cardíacos.
Explanation:
Given:
The cross product is given by
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) =