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bonufazy [111]
3 years ago
7

Does 27/60 -8/60=19/60?

Mathematics
1 answer:
solniwko [45]3 years ago
5 0
Yeah ! It's correct
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Joe's Electric made 15 customer calls last week and 21 calls this week. How many calls must be made next week in order to mainta
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24 is the answer I think
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I need the distance show work please!
Leviafan [203]
For (-7,3) and (5,3) the answer is 12. You add the different points, -7 and 5 because one is negative and one is positive. You do not add 3 and 3 because they are the same.

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3 years ago
Find the surface area of the composite solid.
Illusion [34]

The ends of the rectangle = 10*10*2 = 200

The long sides and bottom of the rectangle = 12*10*3 = 360

Vertical end of the triangle = 5*10 = 50

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3 years ago
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Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia
ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account  the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

              Xa = Xb

     -6 + (3*t) =  7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

         (6.8*t) = 13.4

                  t = 13.4/6.8

                  t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

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This would be the symmetric property!

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