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dolphi86 [110]
3 years ago
14

Please help Asap!!!!!!!!!!!!

Mathematics
2 answers:
Hitman42 [59]3 years ago
8 0
The line segment is 6 units long.
Serhud [2]3 years ago
3 0
All you're doing is counting the units that make up the line, which is 6 units. You could also use the distance formula (\sqrt{(x_{1}-x_{2})^2+( y_{1}+ y_{2})^2  }), but that isn't really necessary since the given line is horizontal.

:)
You might be interested in
What is triangle ABC ?<br><br> (line segments)
Bess [88]
Triangle ABD and ADC are righttriangles so u can use the Pythagorean theorem/triples. the triples are 3/4/5 and 5/12/13. 12 is 4x3, and 15 is 5x3, so BD is 3x3, which is 9. AC is 13, as it is part of the Pythagorean’s triples
7 0
3 years ago
Help me please !!!
Nat2105 [25]

Answer:

For 3x^2+4x+4=0

Discriminant= = -32

The solutions are

(-b+√x)/2a= (-2+2√-2)/3

(-b-√x)/2a= (-2-2√-2)/3

For 3x^2+2x+4=0

Discriminant= -44

The solutions

(-b+√x)/2a= (-1+√-11)/3

(-b-√x)/2a= (-1-√-11)/3

For 9x^2-6x+2=0

Discriminant= -36

The solutions

(-b+√x)/2a= (1+√-1)/3

(-b-√x)/2a= (1-√-1)/3

Step-by-step explanation:

Formula for the discriminant = b²-4ac

let the discriminant be = x for the equations

The solution of the equations

= (-b+√x)/2a and = (-b-√x)/2a

For 3x^2+4x+4=0

Discriminant= 4²-4(3)(4)

Discriminant= 16-48

Discriminant= = -32

The solutions

(-b+√x)/2a =( -4+√-32)/6

(-b+√x)/2a= (-4 +4√-2)/6

(-b+√x)/2a= (-2+2√-2)/3

(-b-√x)/2a =( -4-√-32)/6

(-b-√x)/2a= (-4 -4√-2)/6

(-b-√x)/2a= (-2-2√-2)/3

For 3x^2+2x+4=0

Discriminant= 2²-4(3)(4)

Discriminant= 4-48

Discriminant= -44

The solutions

(-b+√x)/2a =( -2+√-44)/6

(-b+√x)/2a= (-2 +2√-11)/6

(-b+√x)/2a= (-1+√-11)/3

(-b-√x)/2a =( -2-√-44)/6

(-b-√x)/2a= (-2 -2√-11)/6

(-b-√x)/2a= (-1-√-11)/3

For 9x^2-6x+2=0

Discriminant= (-6)²-4(9)(2)

Discriminant= 36 -72

Discriminant= -36

The solutions

(-b+√x)/2a =( 6+√-36)/18

(-b+√x)/2a= (6 +6√-1)/18

(-b+√x)/2a= (1+√-1)/3

(-b-√x)/2a =( 6-√-36)/18

(-b-√x)/2a= (6 -6√-1)/18

(-b-√x)/2a= (1-√-1)/3

6 0
3 years ago
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
Anuta_ua [19.1K]

Answer:

c

Step-by-step explanation:

First, we can transform this into a matrix. The x coefficients will be the first ones for each row, the y coefficients the second column, etc.

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]

Next, we can define a reduced row echelon form matrix as follows:

With the leading entry being the first non zero number in the first row, the leading entry in each row must be 1. Next, there must only be 0s above and below the leading entry. After that, the leading entry of a row must be to the left of the leading entry of the next row. Finally, rows with all zeros should be at the bottom of the matrix.

Because there are 3 rows and we want to solve for 3 variables, making the desired matrix of form

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] for the first three rows and columns. This would make the equation translate to

x= something

y= something

z = something, making it easy to solve for x, y, and z.

Going back to our matrix,

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right] ,

we can start by removing the nonzero values from the first column for rows 2 and 3 to reach the first column of the desired matrix. We can do this by multiplying the first row by -6 and adding it to the second row, as well as multiplying the first row by -1 and adding it to the third row. This results in

\left[\begin{array}{cccc}1&-2&3&-2\\0&14&-16&-36\\0&6&0&-36\end{array}\right]

as our matrix. * Next, we can reach the second column of our desired matrix by first multiplying the second row by (2/14) and adding it to the first row as well as multiplying the second row by (-6/14) and adding it to the third row. This eliminates the nonzero values from all rows in the second column except for the second row. This results in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&14&-16&-36\\0&0&96/14&-288/14\end{array}\right]

After that, to reach the desired second column, we can divide the second row by 14, resulting in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&1&-16/14&-36/14\\0&0&96/14&-288/14\end{array}\right]

Finally, to remove the zeros from all rows in the third column outside of the third row, we can multiply the third row by (16/96) and adding it to the second row as well as multiplying the third row by (-10/96) and adding it to the first row. This results in

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&96/14&-288/14\end{array}\right]

We can then divide the third row by -96/14 to reach the desired third column, making the reduced row echelon form of the matrix

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&1&-3\end{array}\right]

Therefore,

x=-5

y=-6

z=-3

* we could also switch the second and third rows here to make the process a little simpler

3 0
3 years ago
What integer represents a loss of $20
nikklg [1K]
The answer is -20.

The negative part shows the loss and 20 is the amount of money
5 0
3 years ago
Write an equation in slope intercept form with the given information. Goes through points (0,3) and (1,6)
Setler79 [48]

Answer:

below..

Step-by-step explanation:

  • slope = (y - y1)/(x-x1)  = (6 - 3 )/(1-0) = 3 is the slope/ gradient
  • equation of the line =  y -3 = 3(x - 0) => y = 3x + 3
5 0
2 years ago
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