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FinnZ [79.3K]
3 years ago
8

How much larger is the 8 in 8,237 than the 8 in 823.7?

Mathematics
2 answers:
alukav5142 [94]3 years ago
8 0

Answer:Since the 8 in the first number is in the thousands place, its value it 8*1000=8000

8000 is ten times 800.

Step-by-step explanation:

kupik [55]3 years ago
3 0
The 8 in 8,237 is in the thousands place. On the other hand, the 8 in 823.7 is in the hundreds place.

8 • 1,000 = 8,000

8,000 is TEN times 800.

I hope this helped! ~(^v^)~
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Solve the system of equations. <br> 2x+2y+3z=4<br> 5x+3y+5z=5<br> 3x+4y+6z=5
amm1812

Answer:

B. x=3,\ y=20,\ z=-14

Step-by-step explanation:

Consider the system of three equations:

\left\{\begin{array}{l}2x+2y+3z=4\\5x+3y+5z=5\\3x+4y+6z=5\end{array}\right.

Multiply the first equation by 5, the second equation by 2 and subtract them:

\left\{\begin{array}{r}2x+2y+3z=4\\4y+5z=10\\3x+4y+6z=5\end{array}\right.

Multiply the first equation by 3, the second equation by 2 and subtract them:

\left\{\begin{array}{r}2x+2y+3z=4\\4y+5z=10\\-2y-3z=2\end{array}\right.

Multiply the third equation by 2 and add the second and the third equations:

\left\{\begin{array}{r}2x+2y+3z=4\\4y+5z=10\\-z=14\end{array}\right.

Fro mthe third equation

z=-14

Substitute it into the second equation:

4y+5\cdot (-14)=10\\ \\4y=10+70\\ \\ 4y=80\\ \\y=20

Substitute y=20 and z=-14 into the first equation:

2x+2\cdot 20+3\cdot (-14)=4\\ \\2x+40-42=4\\ \\2x-2=4\\ \\2x=6\\ \\x=3

The solution is

x=3,\ y=20,\ z=-14

6 0
3 years ago
The fourth term of an arithmetic sequence is 20, and the 13th term is 65. Find the sum of the first 13 terms
Bess [88]

Answer:

S₁₃ = 455

Step-by-step explanation:

the sum to n terms of an arithmetic is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

We require to find both a₁ and d

The nth term of an arithmetic sequence is

a_{n} = a₁ + (n - 1)d

given a₄ = 20 and a₁₃ = 65 , then

a₁ + 3d = 20 → (1)

a₁ + 12d = 65 → (2)

subtract (1) from (2) term by term to eliminate a₁

9d = 45 ( divide both sides by 9 )

d = 5

substitute d = 5 into (1) and solve for a₁

a₁ + 3(5) = 20

a₁ + 15 = 20 ( subtract 15 from both sides )

a₁ = 5

Then

S₁₃ = \frac{13}{2} [ (2 × 5 ) + (12 × 5 ) }

     = 6.5 (10 + 60)

     = 6.5 × 70

     = 455

7 0
2 years ago
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