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Ne4ueva [31]
3 years ago
5

A 1L bag of Normal Saline must infuse over 6 hours using tubing calibrated to deliver 20gtts/mL. How many drops per minute shoul

d be infused?
Mathematics
1 answer:
sammy [17]3 years ago
4 0

Answer:

total drop per minute is 56

Step-by-step explanation:

Give data:

total capacity of bag 1 L = 1000 ml

Duration of infuse 6 hr

quantity at the time delivered is 20 gtts/ml

Drop per minute can be determined by using following relation

Drop per minute = \frac{1000 ml\times 20 gtts/ml}{6\times 60 min} = 55.55 gtt

therefore total drop per minute is55.55 \approx 56

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Find the function y = f(t) passing through the point (0, 18) with the given first derivative.
monitta

Answer:

\displaystyle y = \frac{t^2}{16} + 18

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
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<u>Algebra I</u>

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<u>Calculus</u>

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Derivative Notation

Antiderivatives - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                                   \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:                                                             \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

\displaystyle \frac{dy}{dt} = \frac{1}{8} t

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

  1. [Derivative] Rewrite:                                                                                         \displaystyle dy = \frac{1}{8} t\ dt
  2. [Equality Property] Integrate both sides:                                                        \displaystyle \int dy = \int {\frac{1}{8} t} \, dt
  3. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 \displaystyle y = \int {\frac{1}{8} t} \, dt
  4. [Right Integral] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle y = \frac{1}{8}\int {t} \, dt
  5. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              \displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C
  6. Multiply:                                                                                                             \displaystyle y = \frac{t^2}{16} + C

<u>Step 3: Find Particular Solution</u>

  1. Substitute in point [Function]:                                                                         \displaystyle 18 = \frac{0^2}{16} + C
  2. Simplify:                                                                                                             \displaystyle 18 = 0 + C
  3. Add:                                                                                                                   \displaystyle 18 = C
  4. Rewrite:                                                                                                             \displaystyle C = 18
  5. Substitute in <em>C</em> [Function]:                                                                                \displaystyle y = \frac{t^2}{16} + 18

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Integration

Book: College Calculus 10e

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