210÷[5+16÷2{6−18÷(7+2)}]−15
2 answers:
Step-by-step explanation:
210 ÷ [ 5 + 16 ÷ 2 { 6 - 18 ÷ ( 7 + 2 ) } ] - 15
210 ÷ [ 5 + 16 ÷ 2 { 6 - 18 ÷ 9 } ] - 15
210 ÷ [ 5 + 16 ÷ 2 { 6 - 2 } ] - 15
210 ÷ [ 5 + 16 ÷ 2 { 4 } ] - 15
210 ÷ [ 5 + 16 ÷ 8 ] - 15
210 ÷ [ 5 + 2 ] - 15
210 ÷ 7 - 15
30 - 15
15
Answer:
Step-by-step explanation:
Just follow the order of operations
210÷[5+16÷2{6−18÷(7+2)}]−15
= 210÷[5+16÷2{6−18÷9}]−15 (calculate 7+2)
= 210÷[5+16÷2{6−2}]−15 (calculate 18÷9)
= 210÷[5+16÷2{4}]−15 (calculate 6-4)
= 210÷[5+16÷2×4]−15
= 210÷[5+8×4]−15 (calculate 16÷2)
= 210÷[5+32]−15 (calculate 8×4)
= 210÷[37]−15 (calculate 5+32)
= 210÷37−15
= 210÷37−(15×37)÷37 (put on the same denominator)
= 210÷37−555÷37 (calculate 15×37)
= (210−555)÷37
=-345÷37 (calculate 210-555)
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Observe the graph below. This graph represents the scenario.
The question is ill formated, the complete question is
In a simulation, a moving object accelerates from rest to 4 meters per second in 2 seconds. For the following three seconds, it increases linearly until it reaches a speed of 10 meters per second. Following three seconds at that speed (acceleration = 0), the item progressively decelerates until it comes to rest two seconds later. Draw the graph of this scenario for 10 seconds?
I'll describe how the graph may show.
It will move diagonally upward from time 0 to 2 seconds until it reaches the y axis at a speed of 4 m/s.
Then, from 2 to 5, the position will move up diagonally until it reaches the y axis at a speed of 10 m/s.
The next 5 to 8 seconds will be horizontal.
After that, it will descend diagonally.
Observe the graph below. This graph represents the scenario.
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