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3 years ago

How can you tell if 2 functions are inverses of eachother

1 answer:

7 0

Remember, if the two graphs are symmetric with respect to the line y = x (mirror images over y = x ), then they are inverse functions. But, we need a way to check without the graphs, because we won't always know what the graphs look like! then f(x) and g(x) are inverse functions.

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For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

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

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

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

8 0

3 years ago

Plisss to na dzisiaj

CaHeK987 [17]

I’m sorry i don’t know the answer but i need to ask another question but i gave up answer a question first sorry

6 0

3 years ago

72.1 written as a fraction

kaheart [24]

It is 72 1/10. This is because 72 is a whole number and point one would be tenths.

5 0

4 years ago

Read 2 more answers

Mateo makes 6 loaves of banana bread. He sells 2/3 of the loves of bread at the bake sale. Which expression represents how many

Lesechka [4]

Answer: 6 * 2/3

Step-by-step explanation:

7 0

4 years ago

Enter the equation of the circle described below. <br> Center (-1,4), radius = sqrt3

slava [35]

The equation of a circle always follows this:

(x - x1) + (y - y1) = r^2

x1 and y1 represent the coordinates as POSITIVE numbers and the r represent the radius

Your answer is (x + 1) + (y - 4) = 3

8 0

3 years ago

Read 2 more answers