Using Gaussian elimination:
x + y + z = 9 / * ( - 2 ) / ( - 1 ) ( we will multiply by - 2 and add to the 2nd 2 x - 3 y + 4 z = 7 equation and multiply by - 1 and add to the 3rd equation )
x - 4 y + 3 z = - 2
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x + y + z = 9
- 5 y + 2 z = - 11 / * ( - 1 )
- 5 y + 2 z = - 11
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x + y + z = 9
- 5 y + 2 z = - 11
0 z = 0
There are infinitely many solutions to the system of equations.
z = t,
- 5 y + 2 t = - 11
5 y = 11 + 2 t
y = ( 11 + 2 t ) / 2
x + ( 11 + 2 t ) / 2 + t = 9 / * 2
2 x + 11 + 2 t + 2 t = 18
2 x = 18 - 11 - 4 t
x = ( 7 - 4 t ) / 2
( x, y , z ) = ( (7 - 4 t ) / 2, ( 11 + 2 t ) / 2, t )
Answer:
14q + 35
Step-by-step explanation:
Step 1: Write expression
7(7q + 5)
Step 2: Distribute 7 to each term
14q + 35
Answer:
All of them
Step-by-step explanation:
According to the ratio test, for a series ∑aₙ:
If lim(n→∞) |aₙ₊₁ / aₙ| < 1, then ∑aₙ converges.
If lim(n→∞) |aₙ₊₁ / aₙ| > 1, then ∑aₙ diverges.
(I) aₙ = 10 / n!
lim(n→∞) |(10 / (n+1)!) / (10 / n!)|
lim(n→∞) |(10 / (n+1)!) × (n! / 10)|
lim(n→∞) |n! / (n+1)!|
lim(n→∞) |1 / (n+1)|
0 < 1
This series converges.
(II) aₙ = n / 2ⁿ
lim(n→∞) |((n+1) / 2ⁿ⁺¹) / (n / 2ⁿ)|
lim(n→∞) |((n+1) / 2ⁿ⁺¹) × (2ⁿ / n)|
lim(n→∞) |(n+1) / (2n)|
1/2 < 1
This series converges.
(III) aₙ = 1 / (2n)!
lim(n→∞) |(1 / (2(n+1))!) / (1 / (2n)!)|
lim(n→∞) |(1 / (2n+2)!) × (2n)! / 1|
lim(n→∞) |(2n)! / (2n+2)!|
lim(n→∞) |1 / ((2n+2)(2n+1))|
0 < 1
This series converges.
Answer:
Step-by-step explanation:
1. 
2. 
3.
4.
5.
6.
7.
605,970 is expanded in this form:
(6 x 100,000) + (0 x 10,000) + (5 x 1,000) + (9 x 100) + (7 x 10) + (0 x1)
each number is multiplied to the corresponding place value.
6 is on the one hundred thousand place value
0 is on the ten thousand place value
5 is on the one thousand place value
9 is on the hundreds place value
7 is on the tens place value
0 is on the ones place value
OR
600,000 + 5,000 + 900 + 70 = 605,970
