I NEED THIS ASAP WHOEVER GETS FIRST GET BRAINLIEST

WILL MARK BRAINLIEST! NEED HELP!

Mice21 [21]

Answer:

The parent graph is translated 2 to the left and up 6 units.

Step-by-step explanation:

The +2 moves the parent 2 units to the left and the + 6 moves it up 6 units.

4 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:

Switch any two rows
Multiply a row by a nonzero constant
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

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%20%5E%208%7D%7B5%20%5E%203%7D" id="TexFormula1" title="\frac{5 ^ 8}{5 ^ 3}" alt="\

Vlada [557]

Step-by-step explanation:

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7 0

3 years ago

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How do you solve this problem ?

svetlana [45]

Answer: 2y124√y12

How to: <u>Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.</u>

Have a great day and stay safe !

3 0

3 years ago

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) Elise has put 5 cans (all of the same size) on her kitchen counter; 2 cans of vegetables, 2 cans of soup , and 1 can of peache

Anvisha [2.4K]

Answer:

0.7 = 70%.

Step-by-step explanation:

There are 5 cans, and she will pick 2, so the number of possibilities that she can pick 2 cans is a combination of 5 choose 2:

C(5,2) = 5!(3!*2!) = 5*4/2 = 10

To find how many possibilities there are with at least 1 can of soup, we can find the number of groups that include no cans of soup, and then see how many possibilities complete the total 10:

There are 3 "no-soup" cans, so the number of possibilities is a combination of 3 choose 2:

C(3,2) = 3!/2! = 3

So, there are 3 possibilities that have no cans of soup, so the number of possibilities that have at least 1 can of soup is 10 - 3 = 7

Then, the probability is 7 / 10 = 0.7 = 70%.

4 0

3 years ago