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3 years ago

DU

Mathematics

1 answer:

Vlad1618 [11]3 years ago

8 0

Answer:

a) W(x) = 2, 000 - 275 x represents the given situation.

b) The cost of the computer after 6 years is $350.

Step-by-step explanation:

Initial worth of the company = $2,000

Depreciation each year = $275

Let, W(x) represent the worth of the computer in x - th year.

So, according to the question:

a) W(x) = 2, 000 - 275 x represents the given situation.

b) Solving W(x) for x = 6, we get

W(6) = 2,000 - 275(6) = 2, 000 - 1,650

= 350

⇒W(6) = $350

Hence, the cost of the computer after 6 years is $350.

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Box A contained 1.67 kg of flour. Box B contained 8600 g of flour. After same amount of

rjkz [21]

Answer:

The amount of flour added to each box was 0.64 kg

Step-by-step explanation:

Let

x ----> the amount of flour added to each box in kg

Remember that

1 kg=1,000 g

8,600 g=8,600/1,000=8.6 kg

we know that

The linear equation that represent this situation is

(8.6+x)=4(1.67+x)

Solve for x

8.6+x=6.68+4x

4x-x=8.6-6.68

3x=1.92

x=0.64 kg

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4 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
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I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

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weeeeeb [17]

Answer:

Plug in x

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Answer:

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Step-by-step explanation:

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3 years ago

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Alona [7]

Answer:

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Step-by-step explanation:

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3 years ago

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