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Shtirlitz [24]
1 year ago
7

the depth of a rain puddle d(t) is given in inches, and t is given in minutes. if the depth is changing with respect to time, wh

ich expression gives the rate of change at which the depth is changing at 3 minutes?
Mathematics
1 answer:
garik1379 [7]1 year ago
7 0

The required expression gives the rate of change will be d'(3) at which the depth is changing at 3 minutes.

Let d(t) represent the depth of the rain puddle at any time t,

⇒ d(t)  ....(i)

The depth is changing with respect to time, which is given in the question

Differentiate equation (i) with respect to t and we get

⇒ δd(t)/δt = d'(t) ....(ii)

This is the rate of change in depth unit time

As per the question, the depth is changing at 3 minutes

Substitute the value of t = 3 in equation (ii), and we get

⇒ d'(3)

Therefore, the required expression gives the rate of change will be d'(3).

To learn more about the differentiation click here:

brainly.com/question/24898810

#SPJ1

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Answer:

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(1/2)600 = 300ml (VODKA)

<em>1) 600cm^3</em>

<em>2) 600ml : 60ml = 10 (GLASSES)</em>

6 0
2 years ago
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Check the picture below.

\bf \left. \qquad  \right.\textit{internal division of a line segment}&#10;\\\\\\&#10;a(-1,2)\qquad b(7,14)\qquad&#10;\qquad 1:3&#10;\\\\\\&#10;\cfrac{aP}{Pb} = \cfrac{1}{3}\implies \cfrac{a}{b} = \cfrac{1}{3}\implies 3a=1b\implies 3(-1,2)=1(7,14)&#10;\\\\&#10;-------------------------------\\\\&#10;{ P=\left(\cfrac{\textit{sum of "x" values}}{r1+r2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{r1+r2}\right)}

\bf -------------------------------\\\\&#10;P=\left(\cfrac{(3\cdot -1)+(1\cdot 7)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 14)}{1+3}\right)&#10;\\\\\\&#10;P=\left( \cfrac{-3+7}{4}~,~\cfrac{6+14}{4} \right)

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3 years ago
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Nezavi [6.7K]
4 because that is what's multiplied throughout
7 0
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Read 2 more answers
If y=a sin 3x+ b cos 3x and <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E%7B2%7D%20y%20%7D%7Bdx%5E%7B2%7D%20%7D" id="TexFormu
Black_prince [1.1K]

Step-by-step explanation:

You have to do first and second derivatives :

y = a sin 3x + b cos 3x

dy/dx = 3a cos 3x - 3b sin 3x

d²y/dx² = - 9a sin 3x - 9b cos 3x

Next, you have to compare it :

let d²y/dx² = ky,

-9a sin 3x - 9b cos 3x = k(a sin 3x + b cos 3x)

-9(a sin 3x + b cos 3x) = k(a sin 3x + b cos 3x)

Therefore, k is equals to 9.

3 0
2 years ago
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