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brilliants [131]

3 years ago

8

Graph the function. g(x)=1/3(x-6)^2+1

1 answer:

sladkih [1.3K]3 years ago

3 0

Step-by-step explanation:

g(x)=1/3(x-6)^2+1

The graph of this function is in the attachment

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Find the slope of the line through each pair of points use the slope formula show your work (-9,-13), (-12,15)

drek231 [11]

Answer:

-28/3

Step-by-step explanation:

The formula for slope is [ y2-y1/x2-x1 ].

15-(-13)/-12-(-9)

28/-3 or -9 1/3

Best of Luck!

7 0

2 years ago

Which transformations are needed to change the parent cosine function to y= 3cos(10(x-pi) )?

denpristay [2]

<span>The parent cosine function can be transformed and translated. So, from the basic function cos(x) we can obtain function acos(bx+c). In our case, a=3- amplitude, b=10- the period change and c=-pi- the phase shift. So, the parent cosine function is mutiplied with 3 (which gives the amplitude of the function, 3*0.5=1.5). The period of the function is changed, and is 2pi/b=2pi/10=pi/5 and the cos(x) is phase shifted for c/b=-pi/10.</span>

6 0

3 years ago

Read 2 more answers

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

AB has one endpoint located at A(-2,

Maru [420]

The other endpoint would be (4,1)

8 0

2 years ago

Estimate the value of 9.9 power of 2 x 1.79 <br> pls some one help me

dybincka [34]

Answer:

The correct answer is 175.4379

Step-by-step explanation:

(9.9)^2 • 1.79

=98.01 • 1.79

=175.4379

3 0

3 years ago