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s344n2d4d5 [400]
2 years ago
14

When you are able to reach a patient by rolling down a window or opening a door, it is known as what type of access?

Health
1 answer:
slega [8]2 years ago
3 0

When you are able to reach a patient by rolling down a window or opening a door, it is known as easy access.

Patient care is offered by emergency medical services (EMS) in a range of situations, from orderly to chaotic. Getting to patients to administer care and removing patients from the site for transfer are both regular issues faced by EMS providers. Starting at dispatch, scene safety is ensured. "EMTs are equipped with the fundamental information and abilities needed to stabilize and transfer patients safely in a variety of situations, from routine medical transports and non-emergencies to life-threatening ones and from easy access to complex situations. Under the direction of doctors, emergency medical technicians work as a part of an extensive EMS response system."

To learn more about patient and easy access here,

brainly.com/question/7322761

#SPJ4

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Anything part of the body that is used to alter the natural voice is referred to as:
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Differentiate the following functions (i) x(1+x)^3​
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Answer:

\displaystyle y' = (1 + x)^2(4x + 1)

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
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  • Function Notation
  • Factoring

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]:                                                                \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                                \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Derivative Rule [Chain Rule]:                                                                                       \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

y = x(1 + x)³

<u>Step 2: Differentiate</u>

  1. Product Rule [Derivative Rule - Chain Rule]:                                                  \displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]
  2. Derivative Property [Addition/Subtraction]:                                                    \displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])
  3. Basic Power Rule:                                                                                             \displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})
  4. Simplify:                                                                                                             \displaystyle y' = (1 + x)^3 + 3x(1 + x)^2
  5. Factor:                                                                                                               \displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]
  6. Combine like terms:                                                                                         \displaystyle y' = (1 + x)^2(4x + 1)

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

Unit: Derivatives

Book: College Calculus 10e

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