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Tresset [83]
2 years ago
8

How do you Round 96,286 to the nearest thousand.

Mathematics
2 answers:
Varvara68 [4.7K]2 years ago
8 0

Answer:

The correct answer is 96,000

Step-by-step explanation:

To round you look at the thousands place and there is a 9 then you look at the number to the right of it which is 2 and it is lower than 5 since it is lower than five it rounds down so the correct answer is 96,000.

HAVE A GOOD DAY!

Leto [7]2 years ago
7 0

Answer:

96,000

Step-by-step explanation:

 If the number on the right of the nearest thousand is lower than 5 the nearest thousand and the numbers to the left of it stay the same, while the numbers on the right turn into 0(zeros) but if the number in the right of the nearest thousand is 5 or grater then the nearest thousand would go up one digit while the numbers on the left of the nearest thousand stay the same, while the numbers on the right of the nearest thousand turn to zeros.

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{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
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There are three possible outcomes that you may encounter when working with these system of equations:


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We are going to try and find values of x, y, and z that will satisfy all three equations at the same time. The following are the equations:

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We are going to use elimination(or addition) method

Step 1: Choose to eliminate any one of the variables from any pair of equations.

In this case it looks like if we multiply the third equation by 4 and  subtracting it from equation 1, it will be fairly simple to eliminate the x term from the first and third equation.

So multiplying Left Hand Side(L.H.S) and Right Hand Side(R.H.S) of 3rd equation with 4 gives us a new equation 4.:

4. 4x-20y-12z = 20      

Subtracting eq. 4 from Eq. 1:

(L.HS) : 4x-2y+5z-(4x-20y-12z) = 18y+17z

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Step 2:  Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.

Similarly if we multiply 3rd equation with 3 and then subtract it from eq. 2 we get:

(L.HS) : 3x+3y+8z-(3x-15y-9z) = 18y+17z

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Step 3:  Solve the remaining system of equations 6 and 5 found in step 2 and 1.

Now if we try to solve equations 5 and 6 for the variables y and z. Subtracting eq 6 from eq. 5 we get:

(L.HS) : 18y+17z-(18y+17z) = 0

(R.HS) : 14-(-11) = 25

0 = 25

which is false, hence no solution exists



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