(75, -80) and (-1, -78). What is the slope of the line that passes through ?

Describe how the graph of each function differs from the graph of f(x)=|x|. Then determine the domain and range.

Brilliant_brown [7]

A. We can compress the graph of the function $f(x)=|x|$ in the y-direction by multiplying the whole function by a constant C, such that $0$ Since $0$ , the graph of the function $g(x)= 0.6|x|$ is obtained from the graph of the function $f(x)=|x|$ by compression in the y-direction.

B. We can stretch the graph of the function $f(x)=|x|$ in the y-direction by multiplying the whole function by a constant C, such that $C>1.$ Since $4>1,$ , the graph of the function $g(x)= 4|x-3|$ is obtained from the graph of the function $f(x)=|x|$ by compression in the y-direction and translation 3 units to the right.

C. We can flip the graph of the function $f(x)=|x|$ upside down by multiplying the whole function by −1, then translate 1 unit to the left and 5 units up. Then we will get the function $g(x)=-|x+1|+5.$

7 0

3 years ago

What is the measure of RMN?

wariber [46]

Answer:

The Answer is: 63°

Step-by-step explanation:

The angle 117° is the key information. Across the line, the angle 117° is part of the 180° across the line. Angle RMN is equal to 180° - 117° = 63°

5 0

3 years ago

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Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

As

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

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7 0

3 years ago

An art gallery wants to display 4 pieces of art in the front window. If there are 8 pieces to choose from, how many distinct dis

Viktor [21]

Answer:

D-70

Step-by-step explanation:

It is a combination C(8,4)

5 0

3 years ago

Who wanna talk to me lol

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers