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4 years ago

5

Michael earns $6.50 an hour plus tips .Last night he received $36.50 in tips and earned a total of $65.75. How many hours did he

work?

2 answers:

Liono4ka [1.6K]4 years ago

8 0

$65.75-$36.50=$29.25
$29.25÷$6.50=4.5 hours
Michael worked 4.5 hours....

Ilya [14]4 years ago

5 0

Michael worked 6 hours

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IgorLugansk [536]

Answer:

3.7m/s^2

Step-by-step explanation:

T is given as 3.27 seconds

L= 1m

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3.27s=2pi(sqrt(1m/g))

3.27s/2pi=(sqrt(1m/g))

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The interquartile range is used as a measure of variability to overcome what difficulty of the range?

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3 years ago

What decimal is equivalent to 122/11?

Lemur [1.5K]

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Sketch the region enclosed by x+y2=12x+y2=12 and x+y=0x+y=0. Decide whether to integrate with respect to xx or yy, and then find

Paladinen [302]

Answer:

$A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}$

Step-by-step explanation:

For this case we have these two functions:

$x+y^2 = 12$ (1)

$x+y=0$ (2)

And as we can see we have the figure attached.

For this case we select the x axis in order to calculate the area.

If we solve y from equation (1) and (2) we got:

$y = \pm \sqrt{12-x}$

$y = -x$

Now we can solve for the intersection points:

$\sqrt{12-x} = -\sqrt{12-x}$

$12-x = -12+x$

$2x=24 , x=12$

$\sqrt{12-x} =-x$

$12-x = x^2$

$x^2 +x -12=0$

$(x+4)*(x-3) =0$

And the solutions are $x =-4, x=3$

So then we have in total 3 intersection point $x=12, x=-4, x=3$

And we can find the area between the two curves separating the total area like this:

$\int_{-4}^3 |\sqrt{12-x} - (-x)| dx +\int_{3}^{12}|-\sqrt{12-x} -\sqrt{12-x}|dx$

$\int_{-4}^3 |\sqrt{12-x} + x| dx +\int_{3}^{12}|-2\sqrt{12-x}|dx$

We can separate the integrals like this:

$\int_{-4}^3 |\sqrt{12-x} dx +\int_{-4}^3 x +2\int_{3}^{12}\sqrt{12-x} dx$

For this integral $\int_{-4}^3 |\sqrt{12-x} dx$ we can use the u substitution with $u = 12-x$ and after apply and solve the integral we got:

$\int_{-4}^3 |\sqrt{12-x} dx =\frac{74}{3}$

The other integral:

$\int_{-4}^3 x dx = \frac{3^2 -(-4)^2}{2} =-\frac{7}{2}$

And for the other integral:

$2\int_{3}^{12}\sqrt{12-x} dx$

We can use the same substitution $u = 12-x$ and after replace and solve the integral we got:

$2\int_{3}^{12}\sqrt{12-x} dx =36$

So then the final area would be given adding the 3 results as following:

$A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}$

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I believe (C), is your answer....Transitive Property of Equality

Hope that helps!!!

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