1 1 1 2
2 -3 1 -11 -2R1 + R2 → R2
-1 2 -1 8 R1 + R3 → R3
1 1 1 2
0 -5 -1 -15 R2 ⇔ R3
0 3 0 10
1 1 1 2
0 3 0 10 -R3
0 -5 -1 -15
1 1 1 2
0 3 0 10 1/3 R2
0 5 1 15
1 1 1 2 -R2 + R1
0 1 0 10/3 -5R2 + R3
0 5 1 15
1 0 1 -4/3
0 1 0 10/3 -R3 + R1
0 0 1 -5/3
1 0 0 1/3
0 1 0 10/3
0 0 1 -5/3
Therefore, x = 1/3, y = 10/3, z = -5/3
Total = principal * (1 + rate/n)^n*years
where "n" is the number of compounding periods per year
Total = 10,000 * (1 + .044/4)^4*2
Total = 10,000 * (1<span>.011</span>)^8
<span><span>Total = 10,000 * 1.0914635699
</span>
</span><span><span>Total = </span>
10,914.64
</span>
Given:
this year's population: 400 students
next year's population: 120% of this year's population
this year's population x rate of next year's population = next year's population
400 students * 120% = 480 students CHOICE D.
There will be 480 students next year.
Answer:
If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day
will be 
.
Step-by-step explanation:
From the question, it is clear that
- If ‘A’ can finish a work in ‘n’ days, then
- we have to determine the part of work finished in 1 day.
So
let '
' be the number of days 
takes to complete the whole work
Let the whole job be denoted as '
'
Thus, the part of work finished in 1 day will be:
Therefore, If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day will be .
Keywords: work, word problem
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Answer:
3
Step-by-step explanation:
It's decreasing by 3 each time, so the next number in the sequence would be 6 - 3, which is 3.
