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Monica [59]
3 years ago
10

Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production of these calcul

ators after t weeks is
dx/dt=5000(1−100/(t+10)2)calculators/week

(Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers’ unfamiliarity with the new techniques.) Find the number of calculators produced from the beginning of the third week to the end of the fourth week.
Mathematics
1 answer:
umka21 [38]3 years ago
6 0

Answer:

The number of calculators is 4871

Step-by-step explanation:

If we integrate dx/dt we get x, which is the number of calculators. To find the number of calculators between the beginning of third week to the end of fourth week (the beginning of fifth week), this integration must be evaluated at t between 3 and 5.

x=\int\limits^5_3 {\frac{dx}{dt}} \, dt =\int\limits^5_3 {5000(1-\frac{100}{(t+10)^2})} \, dt

the result of the integration is:

x=5000(t+\frac{100}{t+10}) to be evaluated between 3 and 5, which is:

x=5000(5+\frac{100}{5+10})-5000(3+\frac{100}{3+10})=\frac{190000}{39}=4871.8

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Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector \vec{v} in some vector space \mathbb R^n we have

\vec{v} = \langle a,b\rangle

This is the component form. I don't like that way. It is probably used in high school, but

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4.

For this question, I think you meant

vectors

\vec{u_1} = (-8, 12)

\vec{u_2}  = (13, 15)

Once

\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}

Considering that the dot product is

\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76

and the norm of \vec{u_1} is ||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}

and the norm of \vec{u_2} is ||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}

Thus,

\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}

\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)

3 0
2 years ago
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El área del paralelogramo con base de 13 centímetros y altura de 8 centímetros es 104 centímetros cuadrados.

Invitamos cordialmente a ver esta pregunta sobre paralelogramos: brainly.com/question/17205536

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