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2 years ago

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A jogger runs around a circular track of radius 55 ft. Let (x,y) be her coordinates, where the origin is the center of the track

. When the jogger's coordinates are (33, 44), her x-coordinate is changing at a rate of 15 ft/s. Find dy/dt.

1 answer:

bagirrra123 [75]2 years ago

6 0

Answer:

11.25 ft/sec

Step-by-step explanation:

Given that a jogger runs around a circular track of radius 55 ft.

. Let (x,y) be her coordinates, where the origin is the center of the track.

Then we know x,y satisfies the equation of the circle as

$x^2+y^2 = 55^2$

Let us differentiate this implicit function with respect to t

$2x \frac{dx}{dt} +2y \frac{dy}{dt} =0$

At this point, dx/dt 15 and x=33. y =44

Substitute to have

$2(33)(15) +2(44) \frac{dy}{dt}\\=0\\\frac{dy}{dt}=11.25$

dy/dt = 11.25 ft /s

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Find the LCM of n^3 t^2 and nt^4.<br><br> A) n 4t^6<br> B) n 3t^6<br> C) n 3t^4<br> D) nt^2

patriot [66]

Answer:

The correct option is C.

Step-by-step explanation:

The least common multiple (LCM) of any two numbers is the smallest number that they both divide evenly into.

The given terms are $n^3t^2$ and $nt^4$ .

The factored form of each term is

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$nt^4=n\times t\times t\times t\times t$

To find the LCM of given numbers, multiply all factors of both terms and common factors of both terms are multiplied once.

$LCM(n^3t^2,nt^4)=n\times n\times n\times t\times t\times t\times t$

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2 years ago

Read 2 more answers

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

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2 years ago

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Illusion [34]

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Let us consider the soda bottles as x and juice bottles as y.

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