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3 years ago

Point W(5, -6) is reflected across the line y = -x. Which of the following is the location of its image, W’.

Mathematics

2 answers:

VashaNatasha [74]3 years ago

8 0

<h3>Answer:</h3>

(6, -5)

<h3>Key Concepts for Reflections:</h3>

if (a, b) is reflected on the x-axis, the image is the point (a, -b)
if (a, b) is reflected on the y-axis, the image is the point (-a, b)
if (a, b) is reflected on the line y = x, the image is the point (b, a)
if (a, b) is reflected on the line y = -x, the image is the point (-b, -a)

(5, -6) --> (-5, 6)

Hope this helps!!

I'd right the key concepts for future reference!

:D

vampirchik [111]3 years ago

3 0

Answer:

(6,-5)

Step-by-step explanation:

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Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$