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3 years ago

The length of a rectangle is 2.5 feet more than its width. If its area is 44.8 square feet,

Mathematics

1 answer:

RideAnS [48]3 years ago

8 0

width = 5.5 feet and

length = 8 feet

Step-by-step explanation:

Given:

Let Width = w

then length = 2.5 + w

Area = 44.8 square feet

Required:

Find length and Width

Formula:

$Area = Length * Width$

Putting values to find length and width

$Area = Length * Width\\44.8=(2.5+w)*w\\44.8=2.5w+w^2\\w^2+2.5w-44.8=0\\Solving\,\,using\,\,quadratic\,\,formula:\\w=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\w=\frac{-(2.5)\pm\sqrt{(2.5)^2-4(1)(-44.8)}}{2(1)}\\w=\frac{-2.5\pm\sqrt{6.25+179.2}}{2}\\w=\frac{-2.5\pm\sqrt{185.45}}{2}\\w=\frac{-2.5\pm13.618}{2}\\w=\frac{-2.5+13.618}{2}\,\,and\,\,w=\frac{-2.5-13.618}{2}\\w=5.5\,\,and\,\,w=-8.1$

So, value of w = 5.5 and -8.1

Since width can't be negative so,

Width = w = 5.5

Length = 2.5+w = 2.5+5.5 = 8

So, width = 5.5 feet and

length = 8 feet

Keywords: Area of rectangle

Learn more about Area of rectangle at:

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Answer:

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Exponential Properties

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

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

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]$

Step 2: Differentiate

Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'$
Derivative Rule [Basic Power Rule]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)$
Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$