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PtichkaEL [24]
3 years ago
11

.The spin on a washing machine takes out 35% of the water in the clothes. The clothes in the washing machine contain 3 pints of

water. How much water is left after a spin?
Mathematics
1 answer:
s2008m [1.1K]3 years ago
4 0

Answer: 1.05 Pints

Step-by-step explanation:

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igor_vitrenko [27]

Answer:

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Step-by-step explanation:

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6 0
2 years ago
Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant
son4ous [18]
The chain rule states that

\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}

Since \ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2), you have

\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}
\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}
\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}

and

\dfrac{\mathrm dx}{\mathrm dt}=\cos t
\dfrac{\mathrm dy}{\mathrm dt}=-\sin t
\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t

Also, since x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t, the derivative is

\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}
\dfrac{\mathrm dw}{\mathrm dt}=\tan t
4 0
2 years ago
6. If the net investment function is given by
Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated  capital for the period.

  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

  • (b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0
2 years ago
Solve for n 20 &gt; -4x - 8
LekaFEV [45]
The answer is X>-7 for your question
5 0
3 years ago
Name two integers whose quotient is -7
Andrei [34K]

Answer:

1 and -7

Step-by-step explanation:

3 0
3 years ago
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