Your 18-year-old patient, Ms. Dex, is a Type 1 diabetic, who arrived in the emergency department in diabetic ketoacidosis. Her blood glucose is over 1050 mg/dL. She received a bolus of isotonic fluid. The physician’s order reads:
Give regular insulin by continuous infusion at the rate of 0.2 units/kg/hour. Leslie weighs 132 lbs.

The medication is available in a solution of 200 units of regular insulin in 100 mL normal saline.

8) How many milliliters (mLs) of the insulin solution will Leslie receive over one hour?

9) The physician calls to inquire about the patient’s condition 1 hour and 30 minutes after the infusion starts. She asks how many units of the insulin have been delivered via the infusion. Calculate how many units of insulin Leslie will receive over 1 hour and 30 minutes (round to the nearest decimal)?

Ms. Dex’s blood glucose level has stabilized. There is a new order for a subcutaneous insulin injection, which reads:
“Check blood glucose levels before meals and at bedtime. For every 4 mg/dl over 140mg/dl, give one unit of Regular Humulin insulin, subcutaneously.”
The nurse knows that the smallest increment that can be drawn up in an insulin syringe is one unit (no partial units).

10) At 5:00 pm, Ms. Dex’s blood glucose level at was measured at 172 mg per deciliter (mg/dl). How many units of insulin will the nurse administer via subcutaneous injection to the patient?

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).