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Dvinal [7]
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?
Mathematics
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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PLEASE HELP ILL GIVE BRAINLIEST
IgorLugansk [536]

Answer:

3.7m/s^2

Step-by-step explanation:

T is given as 3.27 seconds

L= 1m

we have to find g just put values in given formula

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

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

0.520s=(sqrt(1m/g))

Taking square on both side to eliminate square root

0.27085s^2=1m/g

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0.27085 s^2*g=1m

Divide both sides with 0.27085

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4 years ago
The interquartile range is used as a measure of variability to overcome what difficulty of the range?
notsponge [240]
Answers b and c are 'way off (wrong).  Answer d is not negative, and thus could not be the correct answer.  That leaves Answer a:  the range is influenced too much by extreme values.

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3 years ago
What decimal is equivalent to 122/11?
Lemur [1.5K]
\frac{122}{11}=\frac{121+1}{11}=\frac{121}{11}+\frac{1}{11}=11+\frac{1}{11}=11.090909...=11.\overline{09}=11.(09)\\\\\underline{0.090909...}\\1:11\\\underline{0}\\10\\\underline{.\ 0}\\100\\\underline{.\ 99}\\.\ \ 10\\\underline{.\ \ \ 0}\\.\ \ 100\\\underline{.\ \ \ \ 99}\\.\ \ \ \ \ \ 1

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3 years ago
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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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Vesna [10]
I believe (C), is your answer....Transitive Property of Equality



Hope that helps!!!
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