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Morgarella [4.7K]
3 years ago
12

A long-distance runner started on a course at an average of 9 mph half an hour later a second runner begin the same course of an

average of speed of 11 mph how long after the second runner starts the second runner take over the first runner

Mathematics
1 answer:
Firdavs [7]3 years ago
3 0

Answer:

2.25 hours

Step-by-step explanation:

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Identifying angles from three points
USPshnik [31]

we need to identify the angle GFE

We will connect the yellow dots beside the point F which are on the line segments FG and FE

See the following picture:

8 0
1 year ago
GEOMETRY
NeTakaya

the mall if 4 miles east and 5 miles north, which means that the city's center is 4 miles west and 5 miles south to the mall. Truth is 4 miles west and 2 miles south of the city's  center. This means that:

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d. 3 miles is your answer

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8 0
3 years ago
Find the sum and simplify
statuscvo [17]

<u>Answer</u>:

[ ( 2 x^2 - x - 5 ) ] / [ ( x + 2 ) ( x - 3 ) ]

option c). is correct.

<u>Step-by-step explanation</u>:

( x + 1 ) / ( x + 2 ) + ( x - 1 ) / ( x - 3 )

= > [ ( x + 1 ) ( x - 3 ) + ( x - 1 ) ( x + 2 ) ] / [ ( x + 2 ) ( x - 3 ) ]

= > [ ( x^2 - 3 x + x - 3 + x^2 + 2 x - x - 2 ) ] / [ ( x + 2 ) ( x - 3 ) ]

= > [ ( 2 x^2 - x - 5 ) ] / [ ( x + 2 ) ( x - 3 ) ]

Hence we get required answer!

6 0
2 years ago
A tailor needs meters of cloth to make a poncho. How many meters does he need to make 15 ponchos of the same size?
vlabodo [156]

Answer:

15 mister of cloths are needed to make 15 m if 1 puchu is 1 miter

4 0
1 year ago
Read 2 more answers
A cylindrical tank of radius 5 ft and height 9 ft is two-thirds filled with water. Find the work required to pump all the water
Oduvanchick [21]

Answer:

130.826 kilojoules

Step-by-step explanation:

The Work required to pump water = pgV

Where, p = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81ms-¹

V = volume of water

Since, radius r = 5 ft and height h = 9ft

Volume of water in the cylindrical tank = (2/3)πr²h

V = (2/3)π *5²*9 = 471.24 ft³ = 471.24 * 0.0283m³ = 13.336m³

Work required = 1000*9.81*13.336

W = 130.826 KJ

Therefore, the work required to pump water 2/3 of the volume of the cylindrical tank = 130.826 kilojoules.

4 0
3 years ago
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