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schepotkina [342]
2 years ago
7

The recommended dosage for ceclor (cefaclor) is 30mg/kg to be given P.O. for a child weighing 32 lbs. Available: 125mg/5ml. How

much of the solution is needed per day?
Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
4 0

The child's weight in kilogram (kg) is equal to 14.5 kilogram. 2. The safe 24-hour dosage range for this child is 290 mg to 580 mg.

Given that the recommended dosage for ceclor is 30mg/kg to be given P.O. for a child weighing 32 lbs.

In order to determine the weight of the child, we would convert the value of the child's weight in pounds (lbs) to kilogram (kg) as follows:

1 pound = 0.4536 kilogram

32 pound = x kilogram

Cross-multiplying, we have: X = 32 × 0.4536 Child's weight, x = 14.5 kilogram.

Minimum range = 30 × 14.5 = 290 mg to 580 mg

Maximum range = 125 × 14.5 = 580 mg

Therefore, the safe 24-hour dosage range for this child is 290 mg to 580 mg.

Hence based on the calculations, we can infer and logically deduce that yes the dosage ordered safe because the safe single dose range is from 97 mg to 193 mg.

learn more about safe dosage at brainly.com/question/13186642

#SPJ9

Complete Question:

Ceclor (cefaclor) 100 mg p.oh is ordered for a child weighing 32 lbs.

1. What is the child's weight in kilograms (round to tenths)

2. What is safe 24-hour dosage range for this child?

3. Is the dosage ordered safe?

A. Yes

B. No, below the recommended dosage

C. No, it is for above the recommended dosage

D. More information is needed

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marysya [2.9K]

The product is equal to:

\frac{18}{25}(cos(165) + i*sin(165))

<h3>How to solve the product?</h3>

Remember that we can write a complex number in polar form as:

R*e^{i*a} = R*(cos(a) + i*sin(a))

Then the given product:

\frac{6}{5}*(cos(120) + i*sin(120))*\frac{3}{5}*(cos(45) + i*sin(45))

can be rewritten to:

(\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})

Now is easier to solve the product:

(\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})\\\\= \frac{6}{5} *\frac{3}{5} *e^{i*(120 + 45)}\\\\= \frac{18}{25}*e^{i*165}\\\\=  \frac{18}{25}(cos(165) + i*sin(165))

If you want to learn more about complex numbers:

brainly.com/question/10662770

#SPJ1

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