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Anika [276]
3 years ago
11

Mary went to the park with Amy at 2:30 pm. They played for 1 hour 30 min. At what time did the finish playing? Give your answer

in 24-hour clock format and also draw the clock.
Mathematics
2 answers:
puteri [66]3 years ago
6 0

Answer:

4:00 pm

Step-by-step explanation:

deff fn [24]3 years ago
3 0

Answer:

4:00

Step-by-step explanation

First, we add the minutes. 30 + 30 is 60. One hour is 60 minutes so we add 1 when we are adding the hours. 2 + 1 + 1 = 4 so it is 4:00 when they finished playing. I cannot draw the clock because it is not letting me do so.

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Erika paid a self-employment tax last year. She calculated the self-employment tax for different amounts of net earnings and rec
Julli [10]

Answer:

the answer is B

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
HELP
kherson [118]

Answer: For the equation simply substitute 30 in for v and 4 in for s in the equation given, so H=-16t^2+30t+4

Part 2.

If the juggler misses the ball, you wish to know the time that it takes for the ball to travel the four feet from his hand to the ground, so respectively,

4=-16t^2+30t+4 or 0=-16t^2+30t or 0 = -16t+30,

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Step-by-step explanation:

4 0
3 years ago
A phone company charges 50 cents per minute during the daytime and 10 cents per minute at night.
Gala2k [10]

Answer:

You could talk on the phone for 200 minutes at night, or 40 minutes during the day.

For every 5 minutes you reduced your night time talking, you could talk during the day for 1 minute.

Step-by-step explanation:

0.50d + 0.10n < $20

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4 0
3 years ago
4&gt; Solve by using Laplace transform: y'+5y'+4y=0; y(0)=3 y'(o)=o
harina [27]

Answer:

y=3e^{-4t}

Step-by-step explanation:

y''+5y'+4y=0

Applying the Laplace transform:

\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0

With the formulas:

\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)

\mathcal{L}[y']=s\mathcal{L}[y]-y(0)

\mathcal{L}[x]=L

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Solving for L

L(s^2+5s+4)=3s+3

L=\frac{3s+3}{s^2+5s+4}

L=\frac{3(s+1)}{(s+1)(s+4)}

L=\frac3{s+4}

Apply the inverse Laplace transform with this formula:

\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}

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7 0
3 years ago
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