If the patient is less than 70 years old, the loading dose is 80 Howell Units2 per kilogram of DM. One Howell Unit is equivalent to 0.0020 mg of pure heparin. 2. Calculate how many mg of pure heparin is in 80 Howell Units. Do not worry about the kilograms of DM yet. 00.1600mg The loading dose, in mg of heparin for a 5-year-old with a total body weight of 40 lbs., with 10% bodyfat (that is, 90% of that 40 lbs. is lean body mass). You may also be interested to know that 1 kg = 2.20 lbs. Think about the steps you need to go through to find the correct dose. Remember, first you need to find the dosing mass

1) 0.16 mg of pure heparin is in 80 Howell units

2) the loading dose, in mg of heparin for the 5-year-old boy is 2.618 mg

Step-by-step explanation:

Given that;

if a patient is less than 70 years of age then loading dose is 80 Howell units square per Kg of DM,

One Howell unit is equivalent to 0.0020 mg of pure heparin

1)

how many mg of pure heparin is in 80 Howell units.

that will be;

80 × 0.0020 = 0.16 mg

therefore 0.16 mg of pure heparin is in 80 Howell units

2)    

First we find the dosing mass

so dosing mass = body weight of the 5year old boy × 90%

from the question his total body weight of 40 lbs and  1 kg = 2.20 lbs

so

dosing mass = 40lbs × 1kg/2.20lbs × 90/100

dosing mass(kg) = 16.37 kg

Now to calculate the Loading dose we say;

Loading dose = dosing mass × heparin required per kg

Loading dose = 16.36 kg × 0.16 mg heparin per 1kg

loading dose = 2.618 mg heparin

therefore the loading dose, in mg of heparin for the 5-year-old boy is 2.618 mg

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

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