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Alex787 [66]
2 years ago
15

On average, indigo drinks 2/5 of a 10-ounce glass of water in 4/5 of an hour. how much water does she drink, in glasses per hour

? enter your answer as a whole number, proper fraction, or mixed number in simplest form.
Mathematics
1 answer:
yawa3891 [41]2 years ago
5 0

Indigo drinks 5 ounces glasses per hour.

<h3>What is fraction?</h3>

A fraction is used to represent the portion/part of the whole thing. It represents the equal parts of the whole.

As in \frac{4}{5}  hour she drinks =10*\frac{2}{5}

                                     = \frac{20}{5}

                                     =4 ounce

She is drinking 4 ounce in \frac{4}{5}  hours,

= \frac{4}{\frac{4}{5} }

= 4*\frac{5}{4}

=5 ounce in an hour.

She drink 5 ounces glasses per hour.

learn more about fraction here:

brainly.com/question/10354322

#SPJ1

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Answer:

[C]  \displaystyle \frac{-3}{250}

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

<u>Algebra I</u>

  • Terms/Coefficients
  • Factoring
  • Functions
  • Function Notation
  • Conjugations

<u>Calculus</u>

  • Limits
  • Limit Rule [Variable Direct Substitution]:                                                     \displaystyle \lim_{x \to c} x = c
  • Limit Property [Multiplied Constant]:                                                           \displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)
  • Derivatives
  • Definition of a Derivative:                                                                             \displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle g(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

\displaystyle f(x) = \frac{3}{\sqrt{x - 4}}

\displaystyle g(29)

<u>Step 2: Differentiate</u>

  1. Substitute in function [Function g(x)]:                                                           \displaystyle g(x) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{x + h - 4}} - \frac{3}{\sqrt{x - 4}}}{h}
  2. Substitute in <em>x</em> [Function g(x)]:                                                                       \displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{29 + h - 4}} - \frac{3}{\sqrt{29 - 4}}}{h}
  3. Simplify:                                                                                                         \displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{25 + h}} - \frac{3}{5}}{h}
  4. Rewrite:                                                                                                         \displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15}{5\sqrt{25 + h}} - \frac{3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}
  5. [Subtraction] Combine like terms:                                                               \displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15 - 3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}
  6. Factor:                                                                                                           \displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3(5 - \sqrt{25 + h})}{5\sqrt{25 + h}}}{h}
  7. Rewrite:                                                                                                         \displaystyle g(29) = \lim_{h \to 0} \frac{3(5 - \sqrt{25 + h})}{5h\sqrt{25 + h}}
  8. Rewrite [Limit Property - Multiplied Constant]:                                           \displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}}
  9. Root Conjugation:                                                                                         \displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}} \cdot \frac{5 + \sqrt{25 + h}}{5 + \sqrt{25 + h}}
  10. Multiply:                                                                                                         \displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{5h\sqrt{25 + h} + h^2 + 25h}
  11. Factor:                                                                                                           \displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{h(5\sqrt{25 + h} + h + 25)}
  12. Simplify:                                                                                                         \displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + h} + h + 25}
  13. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + 0} + 0 + 25}
  14. Simplify:                                                                                                         \displaystyle g(29) = \frac{3}{5} \cdot \frac{-1}{50}
  15. Multiply:                                                                                                         \displaystyle g(29) = \frac{-3}{250}

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

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