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Ganezh [65]
3 years ago
5

The average high temperature was 86 in August then 74 in September. What the decrease?

Mathematics
2 answers:
Rzqust [24]3 years ago
7 0

Answer:

12

Step-by-step explanation:

86-74=12

You have to subtract the highest by the lowest to find the decrease.

AnnyKZ [126]3 years ago
6 0
86-74= 12

just subtract and you’ll find your answer
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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> </u>

<u>Algebra I</u>

  • Coordinates (x, y)
  • Functions
  • Function Notation
  • Terms/Coefficients
  • Anything to the 0th power is 1
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle y = \sqrt{x - 3}<u />

<u />\displaystyle y' = \frac{1}{2}<u />

<u />

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                                   \displaystyle y = (x - 3)^{\frac{1}{2}}
  2. Chain Rule:                                                                                                        \displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]
  3. Basic Power Rule:                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)
  4. Simplify:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1
  5. Multiply:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}
  6. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                          \displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}
  7. [Derivative] Rewrite [Exponential Rule - Root Rewrite]:                                 \displaystyle y' = \frac{1}{2\sqrt{x - 3}}

<u>Step 3: Solve</u>

<em>Find coordinates</em>

<em />

<em>x-coordinate</em>

  1. Substitute in <em>y'</em> [Derivative]:                                                                             \displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}
  2. [Multiplication Property of Equality] Multiply 2 on both sides:                      \displaystyle 1 = \frac{1}{\sqrt{x - 3}}
  3. [Multiplication Property of Equality] Multiply √(x - 3) on both sides:            \displaystyle \sqrt{x - 3} = 1
  4. [Equality Property] Square both sides:                                                           \displaystyle x - 3 = 1
  5. [Addition Property of Equality] Add 3 on both sides:                                    \displaystyle x = 4

<em>y-coordinate</em>

  1. Substitute in <em>x</em> [Function]:                                                                                \displaystyle y = \sqrt{4 - 3}
  2. [√Radical] Subtract:                                                                                          \displaystyle y = \sqrt{1}
  3. [√Radical] Evaluate:                                                                                         \displaystyle y = 1

∴ Coordinates of Point N is (4, 1).

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

Unit: Derivatives

Book: College Calculus 10e

4 0
2 years ago
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