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Ann [662]
3 years ago
14

Two stabilizing wires extend from the top of a pole to the ground, forming right triangles on

Mathematics
1 answer:
FrozenT [24]3 years ago
6 0

Answer: the height of the pole is 16.1 m.

Step-by-step explanation:

The scenario is illustrated in the attached photo.

x represents the height of the pole.

y represents the distance from the foot of one stabilizing wire to the foot of the pole.

30 - y represents the distance from the foot of the other stabilizing wire to the foot of the pole.

In solving the triangles, we would apply the tangent trigonometric ratio which is expressed as

Tan θ, = opposite side/adjacent side.

Considering triangle ACD,

Tan 60 = x/y

x = ytan60 = y × 1.732

x = 1.732y- - - - - - - - -1

Considering triangle BCD,

Tan 38 = x/(30 - y)

x = (30 - y)tan38 = 0.781(30 - y)

x = 23.43 - 0.781y- - - - - - - - -2

Substituting equation 1 into equation 2, it becomes

1.732y = 23.43 - 0.781y

1.732y + 0.781y = 23.43

2.513y = 23.43

y = 23.43/2.513

y = 9.3

x = 1.732y = 1.732 × 9.3

x = 16.1 m

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Answer:

Max Value: x = 400

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)

<u>Calculus</u>

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  • Integral Property: \int {cf(x)} \, dx = c\int {f(x)} \, dx
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Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = \frac{1}{\sqrt{800-2x} }

<u>Step 2: Identify Variables</u>

<em>Using U-Substitution, we set variables in order to integrate.</em>

u = 800-2x\\du = -2dx

<u>Step 3: Integrate</u>

  1. Define:                                                                                                            \int {f(x)} \, dx
  2. Substitute:                                                                                         \int {\frac{1}{\sqrt{800-2x} } } \, dx
  3. [Integral] Int Property:                                                                                     -\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx
  4. [Integral] U-Sub:                                                                                           -\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du
  5. [Integral] Rewrite:                                                                                          -\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du
  6. [Integral - Evaluate] Reverse Power Rule:                                                 -\frac{1}{2}(2\sqrt{u}) + C
  7. Simplify:                                                                                                         -\sqrt{u} + C
  8. Back-Substitute:                                                                                            -\sqrt{800-2x} + C
  9. Factor:                                                                                                           -\sqrt{-2(x - 400)} + C

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We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

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