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topjm [15]
3 years ago
13

The diagonal of a rectangular swimming pool measures 29 yards. If the length of the pool is 25 yards, which measurement is close

st to the width of the pool?​
Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

Width of the pool = 15 yards

Step-by-step explanation:

Given that:

Length of diagonal of swimming pool = 29 yards

Length of the pool = 25 yards

Let,

w be the width.

This will form a right angled triangle where the diagonal will be hypotenuse.

Using Pythagorean theorem;

(25)^2+(w)^2 = (29)^2\\625+w^2=841\\w^2= 841 - 625\\w^2 = 216

Taking square root on both sides

\sqrt{w^2}=\sqrt{216}\\w=14.70

Rounding off to nearest whole number

w = 15 yards

Hence,

Width of the pool = 15 yards

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The problem is attached, thanks.
NeX [460]

Answer:

\displaystyle \frac{dy}{dx} \bigg| \limit_{(1, 4)} = 2

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

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Coordinates (x, y)
  • Exponential Rule [Root Rewrite]:                                                                 \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}
  • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

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Derivative of a constant is 0

Implicit Differentiation

Basic Power Rule:

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Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle \sqrt{x} - \sqrt{y} = -1

Point (1, 4)

<u>Step 2: Differentiate</u>

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  2. [Implicit Differentiation] Basic Power Rule:                                                 \displaystyle \frac{1}{2}x^{\frac{1}{2} - 1} - \frac{1}{2}y^{\frac{1}{2} - 1}\frac{dy}{dx} = 0
  3. [Implicit Differentiation] Simplify Exponents:                                               \displaystyle \frac{1}{2}x^{\frac{-1}{2}} - \frac{1}{2}y^{\frac{-1}{2}}\frac{dy}{dx} = 0
  4. [Implicit Differentiation] Rewrite [Exponential Rule - Rewrite]:                   \displaystyle \frac{1}{2x^{\frac{1}{2}}} - \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0
  5. [Implicit Differentiation] Isolate <em>y</em> terms:                                                       \displaystyle -\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}
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Unit: Implicit Differentiation

Book: College Calculus 10e

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