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slega [8]
3 years ago
7

Best one gets brainly

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
7 0
I think that it would be a
Brilliant_brown [7]3 years ago
6 0

Answer: The answer is B>3 so if you don't have one with an open circle and going right then the answer is a , but the correct answer would be an open circle to the right or 3

Step-by-step explanation:

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The answer is -3, because the negative sign is outside of the absolute value. If the problem looked like this |-3|, then the answer would be 3. To clarify, your answer is -3.
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Sebastian wanted to order pizzas for $7 each. The delivery charge is $3.50. He has $20.
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Answer:

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What is the correct answer!
AlexFokin [52]

Answer:

The answer is B.

Step-by-step explanation:

Please let me know if you want an explanation for why this is the answer (comment on this). A lot of people don't actually read the explanations, so I wouldn't want to waste my time. However, if you would like it I would be more than happy to type one out for you. Thanks!

6 0
3 years ago
Tom is trying to write 2 3/47 as a decimal
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5 0
3 years ago
{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
lesya692 [45]

There are three possible outcomes that you may encounter when working with these system of equations:


  •    one solution
  •    no solution
  •    infinite solutions

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:

  1. 4x-2y+5z = 6
  2. 3x+3y+8z = 4
  3. x-5y-3z = 5

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

(R.H.S) : 20 - 6 = 14

5. 18y+17z=14

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

(R.H.S) : 4 - 15 = -11

6. 18y+17z = -11

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



3 0
3 years ago
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