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2 years ago

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Miguel and Maria are washing the windows in their home. Together, they can complete the task in 1.5 hours. If Maria can wash win

dows twice as fast as Miguel, how many minutes would it take her to wash them by herself? In your answer, include the equation you used to solve the problem.

2 answers:

snow_tiger [21]2 years ago

8 0

Answer:

2 hours 15 minutes

2.25 hours.

Step-by-step explanation:

Let Maria's time by herself = x minutes

Let Miguel's time by himself = 2x minutes

You have to be careful how you set this equation up. Start with the right. The 1 represents the Job to be done. It takes 90 minutes.

The left hand side = 1/x which is the amount of time Maria (x) takes to do the job by herself. She doesn't do the job by herself.

Miguel helps her, so the portion he does is represented by 1/2x. Because he's helping, her time is cut back; he's doing some of the work.

1/x + 1/(2x) = 1/90 minutes

(2 + 1)/(2x) = 1/90 minutes

3/(2x) = 1/90 minutes

270 = 2x

x = 135 minutes.

It would take her 135 minutes to do the windows alone. That's 2 hours and 15 minutes.

Crazy boy [7]2 years ago

4 0

Answer:For this case, the first thing we must do is define variables.

x: amount of time Miguel uses to complete the task.

y: amount of time Maria uses to complete the task.

We write the system of equations:

x + y = 60

y = (1/2) x

Solving the system we have:

x = 40 minutes

y = 20 minutes

Answer:

it take her to wash them by herself about:

y = 20 minutes

Step-by-step explanation:

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Step-by-step explanation:

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Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

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$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

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