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trapecia [35]
3 years ago
14

Help help help hehelp help

Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0
The answer is C. she should call her supervisor immediately disregarding any chance to forget the customer<span />
Nookie1986 [14]3 years ago
6 0
C, always go with supervisors 

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A family of 2 adults and their child plans to spend the day at the Long beach aquarium. The cost of admission to the Long Beach
Maurinko [17]

Answer:36.75

Step-by-step explanation:

3 0
3 years ago
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Please help ASAP<br> What is the value of m∠A + m∠B?
ICE Princess25 [194]
That would be 90 degrees
7 0
3 years ago
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Angle JKL is 54 degrees. What is the measure of angle MKL? *
hoa [83]

Answer:

MKL = 36°

Step-by-step explanation:

(3x + 6) + (6x + 12) = 54°

3x + 6x + 6 + 12 = 54

9x + 18 = 54

9x = 54 - 18

9x = 36

x = 36/9

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6 0
3 years ago
Erika worked 14 hours last week and 20 hours this week. If she earns $9 per hour, how much did she earn during these two weeks?
romanna [79]
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7 0
3 years ago
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A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s.
Harman [31]
Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
0=-16 t^{2} +122t+99
Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
-99=-16 t^{2} +122t+99
-16 t^{2} +122+198=0

Now we can apply the quadratic formula t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a} where a=-16, b=122, and c=198
t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}
t= \frac{-122+ \sqrt{27556} }{-32} or t= \frac{-122- \sqrt{27556} }{-32}
t= \frac{-122+166}{-32} or t= \frac{-122-166}{-32}
t= \frac{-11}{8} or t=9

Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.
5 0
3 years ago
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