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Rudik [331]
3 years ago
5

The length of a rectangle is 3 ft less than twice the width, and the area of the rectangle is 14 ft2 . find the dimensions of th

e rectangle.
Mathematics
1 answer:
omeli [17]3 years ago
5 0
Given: A= 14 ft²  and  l= 2w-3 
formula for area of a rectangle is A=lw
substitute for what you know and simplify: 
14=(2w-3)w -> w= either -2 or 7/2 but because a width value can not be negative you can eliminate the -2 value. w=7/2 
subsitute width into equation for length and solve
l=2w-3= 2(7/2)-3= 4
width: 3.5 ft
length: 4 ft
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Select the curve generated by the parametric equations. Indicate with an arrow the direction in which the curve is traced as t i
bixtya [17]

Answer:

length of the curve = 8

Step-by-step explanation:

Given parametric equations are x = t + sin(t) and y = cos(t) and given interval is

−π ≤ t ≤ π

Given data the arrow the direction in which the curve is traces means

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The formula of length of the curve is

\int\limits^a_b {\sqrt{\frac{(dx}{dt}) ^{2}+(\frac{dy}{dt}) ^2 } } \, dx

Given limits values are −π ≤ t ≤ π

x = t + sin(t) ...….. (1)

y = cos(t).......(2)

differentiating equation (1)  with respective to 'x'

\frac{dx}{dt} = 1+cost

differentiating equation (2)  with respective to 'y'

\frac{dy}{dt} = -sint

The length of curve is

\int\limits^\pi_\pi  {\sqrt{(1+cost)^{2}+(-sint)^2 } } \, dt

\int\limits^\pi_\pi  \,   {\sqrt{(1+cost)^{2}+2cost+(sint)^2 } } \, dt

on simplification , we get

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\int\limits^\pi_\pi  \,   {\sqrt{(2+2cost } } \, dt

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\sqrt{2} \int\limits^\pi _\pi  {\sqrt{2cos^2\frac{t}{2} } } \, dt

Taking common \sqrt{2} we get ,

\sqrt{2}\sqrt{2}  \int\limits^\pi _\pi ( {\sqrt{cos^2\frac{t}{2} } } \, dt

2(\int\limits^\pi _\pi  {cos\frac{t}{2} } \, dt

2(\frac{sin(\frac{t}{2} }{\frac{t}{2} } )^{\pi } _{-\pi }

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length of the curve is = 4(1+1) = 8

<u>conclusion</u>:-

The arrow of the direction or the length of curve = 8

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3 years ago
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3 years ago
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<h3>Answer:  Choice B. 8/15</h3>

===========================================================

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Answer:

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