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liq [111]
2 years ago
10

A worker was just hit by a forklift the scene is safe and you begin to assess the worker you see that she is conscious and breat

hing normally but obviously in pain what’s is the purpose of doing a secondary survey on this victim
Advanced Placement (AP)
1 answer:
Fiesta28 [93]2 years ago
5 0

The purpose of doing a secondary survey on this victim is to: Look for any potentially life threatening injuries.

<h3>What is secondary survey?</h3>

Secondary survey are body survey carried out on a victim so as to detect any form of life threatening injuries and this survey is carried out by examining the whole body of the victim.

Life-threatening injuries are injuries that can lead to death if not diagnosed and treated.

Based on the information given carrying out secondary survey on this victim as the victim may have apparent injury.

Hence, secondary survey is important as it will help to detect any  life threatening injuries that may not be immediately visible.

Example of  life-threatening injuries includes:

  • Head injury
  • Skull fractures
  • Neck injury
  • Spinal cord injury etc

Inconclusion the purpose of doing a secondary survey on this victim is to: Look for any potentially life threatening injuries.

Learn more about secondary survey here:brainly.com/question/1695365

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The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
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Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

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