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Nata [24]
3 years ago
8

An airplane takes 4 hours to travel a distance of 2600 miles with the wind. The return trip takes 5 hours against the wind. Find

the speed of the plane in still air and the speed of the wind.
Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0
2,600/4 =650
650•5 =3,250


I found the unit rate of the 2,600 by dividing the per hour and then times it by the 5 hours

Hopefully that was what you were looking for
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Answer:

A. x + 5

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \frac{x^2 - 25}{x - 5}

<u>Step 2: Simplify</u>

  1. Factor:                                                                                                               \displaystyle \frac{(x - 5)(x + 5)}{x - 5}
  2. Divide:                                                                                                            \displaystyle x + 5
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Answer:

Step-by-step explanation:

Given expression is,

\text{cot}A=\frac{1}{2}(\text{cot}\frac{A}{2}-\text{tan}\frac{A}{2})

To prove this identity we will take the right side of the identity,

\frac{1}{2}(\text{cot}\frac{A}{2}-\text{tan}\frac{A}{2})=\frac{1}{2}(\frac{1}{\text{tan}\frac{A}{2}}-tan\frac{A}{2})

                         =\frac{1}{2}(\frac{1-\text{tan}^2\frac{A}{2}}{tan\frac{A}{2}})

                         =\frac{1}{2}[\frac{2(1-\text{tan}^2\frac{A}{2})}{2tan\frac{A}{2}}]

                         =\frac{1}{2}(\frac{2}{\text{tan}A} ) [Since \text{tan}A=\frac{2\text{tan}\frac{A}{2}}{1-\text{tan}^2\frac{A}{2}}]

                         = cot A

Hence right side of the equation is equal to the left side of the equation.

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