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Nata [24]
2 years ago
15

1. A motorboat travels 196 meters in 20 seconds. What is the motorboat's unit rate of speed?

Mathematics
2 answers:
Grace [21]2 years ago
8 0

Answer:

9/4/5m/s

speed is equal to distance all over time. So196m/20s=9/4/5m/s

dezoksy [38]2 years ago
3 0

Answer:

196: 9.8

20: 1

x= 9.8 meters in 1 second

Step-by-step explanation:

This can be determined by following writing equal ratios that represent a word problem. Then, the denominator must be divided from the numerator to get the unit rate/ meters per second.

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Please help me I need help
dsp73

Answer:

i think it would be 87% i have been working on ths for a mnet and it got confusing for a sec i may be rong but i am pretty shure that this is the answer

btw i cant spell

Step-by-step explanation:

8 0
3 years ago
Q.1 Simplify :-<br><br>(i) (- 625 ) + 925 + 100 + (-200)​
Dmitrij [34]
825 - 625i

Hope it’s right
Best luck with your studying
3 0
2 years ago
Rory's battery was full at 6:00 a.M, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three o
Taya2010 [7]

Complete question:

Rory records the percentage of battery life remaining on his phone throughout a day. The graph represents the percentage of battery life remaining after a certain number of hours.

Rory’s battery was full at 6:00 a.m, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three options.

12:00 p.m.

3:00 p.m.

5:00 p.m.

8:00 p.m.

11:00 p.m.

The graph is attached

Answer:

3:00 p.m

5:00 p.m

11:00 p.m

Step-by-step explanation:

From the graph, there are rise times, fall times and there are times the movement is steady (no rise or fall).

The rise time represents the time the phone was plugged. The fall time represents when the charger has been unplugged so the battery level starts depreciating. When it is constant, it means the battery level is at 100 but the charger is still connected to the phone.

We are told at initial condition, time was 6 a.m (t = 0). To get the exact time, we are to add the initial condition, i.e 6

From the graph, the times the phone could have been plugged are 8:00 to 12:00 hrs and 16:00 to 20:00 hrs.

Converting from 24 hr to 12 hr time, we have:

6 + 8hrs = 14:00 = 2:00 p.m

6 + 9hrs = 15:00 = 3:00 p.m

6 + 10 hrs = 16:00 = 4:00 p.m

6 + 11 hrs = 17:00 = 5:00 p.m

6 + 12 hrs = 18:00 = 6:00 p.m

6 + 16 hrs = 22:00 = 10:00 p.m

6 + 17 hrs = 23:00 = 11:00 p.m

6 + 18 hrs = 00:00 = 12:00 am

6 + 19 = 1 : 00 = 1:00 a.m

6 + 20 = 2:00 = 2:00 a.m

From the options given in the question, we have:

3:00 p.m; 5:00 p.m; 11:00 p.m

Therefore, times could Rory's phone have been plugged into the charger are:

3:00 p.m

5:00 p.m

11:00 p.m

5 0
3 years ago
Please help me <br>ineed it now ​
yanalaym [24]

Answer:

look at explanation

Step-by-step explanation:

To solve all of these questions you just need to scale. While scaling you can divide or multiply.

1. \frac{180 km}{4 hrs} (divide by 4 to get unit rate)

     unit rate = \frac{45km}{1hr}

2. If it takes one hour to travel 40 miles, then we don't have to scale or anything. We know \frac{40 miles}{1 hr}, the answer is...

"It'll only take one hour to travel 40 miles. This is correct because the unit rate is \frac{40 miles}{1 hr}. "

3. Knowing that the constant speed of the plane is \frac{800 km}{1 hour\\}, we can scale then add half of the unit rate to get our answer.

\frac{800 km}{1 hour\\} x 3 = \frac{2400 km}{3 hrs}   800 x 3 = 2400

then add half of unit rate which is \frac{400 km}{30 mins}  \sqrt[2]{800} = 400

so, our answer is \frac{2800 km}{3.5 hours}

4.  \frac{3km}{30 mins} <--- ( divided by 2) \frac{6 km}{1 hr} ( multiplied x 2.5) ---> \frac{15 km}{2.5 hrs}

To get an easier way to find the answer, if possible you can scale back farther than the 1 hour mark.

Elmer would take 2 and a half hours to ride 15 km.

5. The boys speed is \frac{4 km}{1 hr}. All we need to do is divide by 2. \sqrt[2]{8} = 4

3 0
3 years ago
Will mark bianleast
Sauron [17]

Answer:

The mean is 7.975.

Step-by-step explanation:

5 0
3 years ago
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