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rodikova [14]
3 years ago
6

I need help fast plz​

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Answer:

what do you need help with

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Which of the following graphs represents a function with a domain of (-∞, ∞) and a range of (-∞, ∞)?
umka21 [38]
The negative infinity for the x coordinate states that the graph should move to the bottom and the y coordinate is positive infinity so that the graph goes up

the first graph is your answer
6 0
3 years ago
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Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI
cricket20 [7]

Answer:

The system has infinitely many solutions

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

  1. Switch any two rows
  2. Multiply a row by a nonzero constant
  3. Add one row to another

To solve the following system

\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}

Step 1: Transform the augmented matrix to the reduced row echelon form

\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

which corresponds to the system

\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}

The system has infinitely many solutions.

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

7 0
3 years ago
Question 7 E Worth 5 points)
Zolol [24]

Answer:

45!

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Find the gradients of line a and b
gayaneshka [121]

Answer:

Gradient of A: 2

Gradient of B: -1

Step-by-step explanation:

Gradient = change in y/change in x

✔️Gradient of A using two points on line A, (2, 5) and (0, 1):

Gradient = (1 - 5)/(0 - 2) = -4/-2

Simplify

Gradient of A = 2

✔️Gradient of B using two points on line B, (0, 5) and (5, 0):

Gradient = (0 - 5)/(5 - 0) = -5/5

Simplify

Gradient of B = -1

5 0
2 years ago
Estimate the slope of the tangent line of the function y(t)=\sqrt{(3t+1} at the point t=2.
Korvikt [17]

Answer: 0.567

Step-by-step explanation:

y(t)=\sqrt{3t+1} \\\\y'(t)=\dfrac{3}{2\sqrt{3t+1} } \\\\y'(2)=\dfrac{3}{2\sqrt{3*2+1} } =\dfrac{3\sqrt{7} }{14} =0,56694670951384084082177480435127\approx{0.567}

7 0
3 years ago
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