Dopamine is available as 400 mg in 250 ml of d5w. a 2 year old weighing 12 kg is receiving 10 mcg/kg/min. how many hours will the infusion last?

55.5556 hours.

Step-by-step explanation:

Let's solve the problem.

The amount of dopamine rate applied to a person is based on the formula: 10mcg/kg/min. Such relation can be express as follows:

(10mcg/kg/min)=

(10mcg/kg)*(1/min)

Now by multiplying by the weight (12 kg) of the 2 years old person, we have:

(10mcg/kg)*(1/min)*(12kg)=

(10mcg*12kg/kg)*(1/min)=

(120mcg)*(1/min)=

120mcg/min, which is the rate of dopamine infusion, which can be express as:

(120mcg/min)*(60min/1hour)=

(120mcg*60min)/(1hour*1min)=

7200mcg/hour=

1hour/7200mcg, which means that for each hour, 7200mcg dopamine are infused.  

Because the D5W product has 400 mg of dopamine, then we need to convert 400 mg to X mcg of dopamine in order to use the previous obtained rate. This means:

Because 1mcg=0.001mg then:

(400mg)*(1mcg/0.001mg)=

(400mg*1mcg)/(0.001mg)=

400000mcg, which is the amount of dopamine in D5W.

Now, using the amount of dopamine in D5W and the applied rate we have:

(rate)*(total amount of dopamine)=hours of infusion

(1hour/7200mcg)*(400000mcg)=hours of infusion

(1hour*400000mcg)/(7200mcg)=hours of infusion

(55.5556 hours) =hours of infusion

In conclusion, the infusion will last 55.5556 hours.

  q

step-by-step explanation:

each line from a vertex to the midpoint of the opposite side is called a median. the point where the medians intersect, point q, is the centroid.

the centroid of δabc is point q.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.