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kiruha [24]
3 years ago
10

Give the simple formula for the nth term of the following arithmetic sequence. Your answer will be of the form an + b.

Mathematics
1 answer:
klasskru [66]3 years ago
6 0
14, 18, 22, 26, 30, . . . = 14, 14 + 4, 14 + 8, 14 + 12, 14 + 16, . . . = 14, 14 + 1(4), 14 + 2(4), 14 + 3(4), 14 + 4(4), . . . = 14 + 4(n - 1) = 14 + 4n - 4 = 4n + 10
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Write -0.06 as a fraction
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-\frac{6}{100}

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according to "Black Hole Beginnings," the destruction of some stars results in inbursts of light. an intense gravitational pull.
Colt1911 [192]

According to "Black Hole Beginnings," the destruction of some stars results in an an intense gravitational pull.

To answer the question, we need to know what Black Holes are.

<h3>What are Black Holes?</h3>

Blackholes are regions of space in which no matter can escape

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brainly.com/question/6037502

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3 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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