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Nat2105 [25]
3 years ago
10

Can anyone help with this and explain...thanks so much!​

Mathematics
2 answers:
Oxana [17]3 years ago
5 0

Answer:

A" (-6, 3)

B" (-9, 0)

C" (-5, -1)

Step-by-Step explination:

You just add 4 to x then make it negative since it is reflecting over the y-axis(the vertical line). Then add 1 to y but it stays the same after that.

Sorry kind of bad at explaining but thats the best i can do for this problem, hoped it helped...

oksano4ka [1.4K]3 years ago
4 0

This is for reflection across the y-axis

A (-2,2)

B (-5,-1)

C (-1,-2)

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MAXImum [283]

Answer:

\displaystyle y'(27) = \frac{1}{27}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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  6. Addition
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  • Left to Right<u> </u>

<u>Algebra I</u>

  • Exponential Rule [Root Rewrite]: \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}
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Derivative Notation

Basic Power Rule

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Step-by-step explanation:

<u>Step 1: Define</u>

y = ∛x

x = 27

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                               \displaystyle y = x^{\frac{1}{3}}
  2. Basic Power Rule:                                                                                         \displaystyle y' = \frac{1}{3}x^{\frac{1}{3} - 1}
  3. Simplify:                                                                                                         \displaystyle y' = \frac{1}{3}x^{-\frac{2}{3}}
  4. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle y' = \frac{1}{3x^{\frac{2}{3}}}

<u>Step 3: Solve</u>

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  2. Evaluate exponents:                                                                                     \displaystyle y'(27) = \frac{1}{3(9)}
  3. Multiply:                                                                                                         \displaystyle y'(27) = \frac{1}{27}

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Unit: Derivatives

Book: College Calculus 10e

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Cos ø = - 46 / 140.72

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