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Deffense [45]
3 years ago
9

Nila's soccer practice started at 8:30 A.M. on Saturday morning. The team practiced dribbling for 1 hour and practiced shooting

for 15 minutes. Then they played a scrimmage game for 1 hour and 15 minutes before practice ended. What time was it when Nila's soccer practice ended?
Mathematics
1 answer:
atroni [7]3 years ago
3 0

Answer:

11 pm

Step-by-step explanation:

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This is easy, if adding the first number to the second number you get 70.2 then if you want to find the middle number you will have to subtract 32.34 from 70.2 to get 37.86, now add that and 32.34 to check if it gives you 70.2.... When you add them you indeed get 70.2 so your work is correct meaning your middle number is 37.86

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3 years ago
I need help please.
Norma-Jean [14]

Answer:

No

Step-by-step explanation:

7 0
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What is 27/64 in simplest form
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The fraction is already in simplest form!:)
4 0
3 years ago
Use a paragraph, flow chart, or two-column proof to prove the angle congruency. Given: ∠ CXY ≅ ∠ BXY
Llana [10]

Answer:

See explanation

Step-by-step explanation:

Consider triangles CAX and BAX. In these triangles,

  • \overline{AC}\cong \overline{AB} - given
  • \angle CAX\cong \angle BAX - given
  • \overline{AX}\cong \overline{AX} - reflexive property

By SAS postulate, \triangle CAX\cong \triangle BAX

Congruent triangles have conruent corresponding parts. So,

\overline{CX}\cong \overline{BX}

Consider triangles CXY and BXY. In these triangles,

  • \overline{CX}\cong \overline{BX} - proven
  • \angle CXY\cong \angle BXY - given
  • \overline{XY}\cong \overline{XY} - reflexive property

By SAS postulate, \triangle CXY\cong \triangle BXY

Congruent triangles have conruent corresponding parts. So,

\angle XCY\cong \angle XBY

5 0
3 years ago
Karen claps her hand and hears the echo from a distant wall 0.113 s later. How far away is the wall? The speed of sound in air i
insens350 [35]

Answer:

19.3795 m

Step-by-step explanation:

If sound travels at 343 m/s and it took Karen 0.113 s to hear the echo from the wall, the distance travelled by the sound is:

D=343\frac{m}{s} *0.113\ s\\D= 38.759\ m

Note that the distance calculated above is the distance travelled from Karen to the wall and then back from the wall to Karen. Therefore, the distance between Karen and the wall is:

d=\frac{38.759}{2}\\d=19.3795\ m

The wall is 19.3795 m away from Karen.

3 0
3 years ago
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