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guapka [62]
3 years ago
5

A rectangular yard has area 96 square feet. If the width of the yard is 4 ft less than length, what is the perimeter,in feet, of

the yard?
Mathematics
1 answer:
marusya05 [52]3 years ago
6 0
Area of the rectangular yard = 96 square feet
Let us assume the length of the yard = x feet
Width of the yard = x - 4 feet
Then
Area of the yard = length * width
96 = x(x - 4)
96 = x^2 - 4x
x^2 - 4x - 96 = 0
x^2 - 12x + 8x - 96 = 0
x(x - 12) +8(x - 12) = 0
(x - 12)(x + 8) = 0
As the value of length cannot be negative so
x - 12 = 0
x = 12
So
Length of the yard = 12 feet
Width of the yard = x - 4 feet
                             = 12 - 4 feet
                             = 8 feet
Then
Perimeter of the yard = 2 (length + width)
                                   = 2(12 + 8)
                                   = 2 * 20 feet
                                   = 40 feet.
So the perimeter of the yard  is 40 feet.
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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
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cos\ \varnothing = \frac{-7}{10}\sqrt 2

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Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

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r = \sqrt{x^2 + y^2

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x = -7 and y = -1

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r = \sqrt{50

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sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

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sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

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sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

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