Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.
xy = 6
(a) Find dy/dt, given x = 4 and dx/dt = 11.
dy/dt =

(b) Find dx/dt, given x = 1 and dy/dt = –9.
dx/dt =

Answers

Answer 1

Answer:

A)

[tex]\displaystyle \frac{dy}{dt}=-\frac{33}{8}[/tex]

B)

[tex]\displaystyle \frac{dx}{dt}=\frac{3}{2}[/tex]

Step-by-step explanation:

x and y are differentiable functions of t, and we are given the equation:

[tex]xy=6[/tex]

First, let's differentiate both sides of the equation with respect to t. So:

[tex]\displaystyle \frac{d}{dt}\left[xy\right]=\frac{d}{dt}[6][/tex]

By the Product Rule and rewriting:

[tex]\displaystyle \frac{d}{dt}[x(t)]y+x\frac{d}{dt}[y(t)]=0[/tex]

Therefore:

[tex]\displaystyle y\frac{dx}{dt}+x\frac{dy}{dt}=0[/tex]

A)

We want to find dy/dt when x = 4 and dx/dt = 11.

Using our original equation, find y when x = 4:

[tex]\displaystyle (4)y=6\Rightarrow y=\frac{3}{2}[/tex]

Therefore:

[tex]\displaystyle \frac{3}{2}\left(11\right)+(4)\frac{dy}{dt}=0[/tex]

Solve for dy/dt:

[tex]\displaystyle \frac{dy}{dt}=-\frac{33}{8}[/tex]

B)

We want to find dx/dt when x = 1 and dy/dt = -9.

Again, using our original equation, find y when x = 1:

[tex](1)y=6\Rightarrow y=6[/tex]

Therefore:

[tex]\displaystyle (6)\frac{dx}{dt}+(1)\left(-9)=0[/tex]

Solve for dx/dt:

[tex]\displaystyle \frac{dx}{dt}=\frac{3}{2}[/tex]


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Answer: 4 gallons per day

Step-by-step explanation:

From the question, we are informed that the Snyder family uses up a 12 -gallon jug of milk every 3 days.

The rate at which they drink milk will be gotten by dividing the gallons of milk they drink by the number of days. This will be:

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Answers

Answer:

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Step-by-step explanation:

To solve this question, we will use the formula for calculating the amount formula as shown;

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Answers

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Answers

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Step-by-step explanation:

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Answers

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answer

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Step-by-step explanation:

set it equal to 180 because the total sum of the interior angles of a triangle is equal to 180

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2x=100

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now plug it in

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Answers

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