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r-ruslan [8.4K]
3 years ago
11

From start to finish, a machine can fill a bottle of water, seal it, label it, and pack it in 8.6 seconds. About how many bottle

s of water are packed in 480 seconds?
Mathematics
2 answers:
bearhunter [10]3 years ago
4 0

Answer:

56!

Step-by-step explanation:

To find the number of bottles that can be packed in 480 seconds, you need to divide the total amount of time, 480 seconds, by 8.6, the amount of time it takes to pack each bottle.

480/8.6=55.8139534884, which rounds up to 56!

lutik1710 [3]3 years ago
4 0

Answer:

56 bottles of water are packed in 480 seconds.

Step-by-step explanation:

8.6 seconds = 1 bottle

480 seconds = ?

480/8.6

= 55.813 approx. 56 bottles

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We start by setting up the system

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Using one of the two equations we can derive the correspondent y value:

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We choose the other two pairs of lines to find the other vertices:

\begin{cases}y=-x+2\\y=x-2 \end{cases} \iff -x+2=x-2 \iff x=2 \implies y = 0

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So, the three vertices are (1, 1), (2, 0), (-1, -3).

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a) The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }.

b) The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}.

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}.

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable (y, t) in each side of the identity:

6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t (1)

6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C

Where C is the integration constant.

By table of integrals we find the solution for each integral:

-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C

If we know that x = 0 and y = 1<em>, </em>then the integration constant is C = -2.

The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }. \blacksquare

b) In this case we need to solve a first order ordinary differential equation of the following form:

\frac{dy}{dx} + p(x) \cdot y = q(x) (2)

Where:

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Hence, the ordinary differential equation is equivalent to this form:

\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x} (3)

The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}. \blacksquare

The solution for (2) is presented below:

y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C (4)

Where C is the integration constant.

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y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C

y = x\int {x} \, dx + x\int\, dx + C

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y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}. \blacksquare

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

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