Answer:
x <u><</u> 4 or 4 <u>-</u> x
Step-by-step explanation:
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Answer:
q = 0
Step-by-step explanation:
8q + 12 = 4( 3 + 3q )
Distributive property
8q + 12 = 4 ( 3 + 3q )
8q + 12 = 12 + 12q
Inverse operations
8q + 12 = 12 + 12q
-8q -8q
12 = 12 + 4q
-12 -12
0 = q
q = 0
First we'll consider the floor size.
In a 10 by 12 room the longest pipe that will fit lying on the floor equals the hypotenuse of a 10 by 12 triangle:
hypotenuse² = 10² + 12²
hypotenuse² = 100 + 144
hypotenuse² = 244
hypotenuse =
<span>
<span>
<span>
15.62 feet
Now we have to consider the third dimension (height = 8 feet).
The two sides would be 15.62 and 8
</span></span></span>hypotenuse² =
15.62² + 8²
hypotenuse² =
244 + 64 = 308
hypotenuse = 17.55
So, the longest pipe that would fit in that room would be 17.55 feet long.
Answer:
conduction
Step-by-step explanation:
Radiation transfers heat to the pavement through empty space. Because the pavement is black, it absorbs more heat. ... Because the pavement is darker so it absorbs more heat.
Answer:
If we arrange the talks from the lowest starting time to the highest ending time we get total of 11 talks.
using algorithm 7 we get answer (1) - (3) - (6) - (9) the largest number of talks scheduled.
Step-by-step explanation:
arranging the talks from lowest starting time to the highest ending time.
thus,
- 9:00 a.m. and 9:45 a.m.
- 9:30 a.m. and 10:00 a.m.
- 9:50 a.m. and 10:15 a.m.
- 10:00 a.m. and 10:30 a.m.
- 10:10 a.m. and 10:25 a.m.
- 10:30 a.m. and 10:55 a.m.
- 10:15 a.m. and 10:45 a.m.
- 10:30 a.m. and 11:00 a.m.
- 10:45 a.m. and 11:30 a.m.
- 10:55 a.m. and 11:25 a.m.
- 11:00 a.m. and 11:15 a.m.
we start from the earliest time as
9:00 a.m. and 9:45 a.m which is (1).
After the talk is finished we pick the nearest time for another talk which starts at
9:50 a.m. and 10:15 a.m which is (3).
After this talk we again pick the nearest time for another talk which becomes
10:30 a.m. and 10:55 a.m which is (6).
and lastly
10:45 a.m. and 11:30 a.m which is (9).
Note: we didn't choose other times because we cannot talk at 2 or 3 places at the same time. so we pick another when one talk is finished.
thus the answer is (1) - (3) - (6) - (9) as the largest number of talks scheduled.
