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algol13
2 years ago
15

The rules for a race require that all runners start at $A$, touch any part of the 1200-meter wall, and stop at $B$. What is the

number of meters in the minimum distance a participant must run
Mathematics
1 answer:
mixer [17]2 years ago
5 0

Let the place that they touch be x meters from the 300 m wall

Then the distance from the 500 m wall will be 1200 - x

Then....the total distance, D, can be expressed asD = sqrt [ x^2 + 300^2 ] + sqrt [ (1200 - x)^2 + 500^2 ]

This can be solved with Calculus, but it's a little messy

The graph shows that the minimum distance is achieved when x = 450m

And the minimum distance  ≈  1442 m

By the way......we can show that this is true if

arctan [ 300 / 450]   =  arctan [ 500 / 750 ]  ......and this is true  !!!!

Learn more about Distance:

brainly.com/question/12356025

#SPJ4

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Solve. Round to the nearest tenth.<br> 6<br> 2-12<br> 19<br> 2x-2<br> HELP ASAP!!
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Answer:

\displaystyle x = \frac{216}{7}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle \frac{6}{19} = \frac{x - 12}{2x - 2}<u />

<u />

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Cross-multiply:                     \displaystyle 6(2x - 2) = 19(x - 12)
  2. Distribute:                             \displaystyle 12x - 12 = 19x - 228
  3. Isolate <em>x</em> terms:                     \displaystyle -12 = 7x - 228
  4. Isolate <em>x</em> term:                      \displaystyle 216 = 7x
  5. Isolate <em>x</em>:                               \displaystyle \frac{216}{7} = x
  6. Rewrite:                                \displaystyle x = \frac{216}{7}
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