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3 years ago

What's the area of this rectangle?

Mathematics

1 answer:

Bess [88]3 years ago

7 0

well, the assumption is that is a rectangle, namely it has two equal pairs, so we can just find the length of one of the pairs to get the dimensions.

hmmmm let's say let's get the length of the segment at (-1,-3), (1,3) for its length

and

the length of the segment at (-1, -3), (-4, -2) for its width

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{length}{L}=\sqrt{[1-(-1)]^2+[3-(-3)]^2}\implies L=\sqrt{(1+1)^2+(3+3)^2} \\\\\\ L=\sqrt{4+36}\implies L=\sqrt{40} \\\\[-0.35em] ~\dotfill$

$\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{width}{w}=\sqrt{[-4-(-1)]^2+[-2-(-3)]^2}\implies w=\sqrt{(-4+1)^2+(-2+3)^2} \\\\\\ w=\sqrt{9+1}\implies w=\sqrt{10} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{A=Lw}\implies \sqrt{40}\cdot \sqrt{10}\implies \sqrt{400}\implies \boxed{20}$

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a) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1$

General Formulas and Concepts:

Pre-Calculus

Unit Circle

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

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

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

Trigonometric Differentiation

Logarithmic Differentiation

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)$

Step 2: Differentiate

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}$
Basic Power Rule [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}$

Step 3: Find

Substitute in x = 0 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = 0} = \frac{0 + 1}{(0 - 1)(0^2 + 1)}$
Evaluate: $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

Step 1: Define

Identify

$\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)$

Step 2: Differentiate

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}$
Trigonometric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}$

Step 3: Find

Substitute in x = π/2 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = \frac{-[cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) - 1]}{[sin(\frac{\pi}{2}) + 1][cos(\frac{\pi}{2}) - 1]}$
Evaluate [Unit Circle]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1$

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

Unit: Differentiation

Book: College Calculus 10e

4 0

3 years ago