Which of the f -values satisfy the following inequality?

I have no idea what the answers are

lara [203]

Hello from MrBillDoesMath!

Answer:

Top line: y = (2/3)x + 2

Bottom line: y = (2/3)x -1

Discussion:

The graph provided is hard to read but I did the best I could.

The top line appears to pass through the points (0,2) and (-3,0)

For this line

m = change y /change x = (0-2)/(-3-0) = -2/-3 = +2/3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,2) set x = 0, y= 2 in y = (2/3)x + b =>

2 = (2/3) 0 + b => b = 2

Therefore y = (2/3)x + 2

The bottom line appears to pass through the points (0,-1) and (3,1)

For this line

m = change y /change x = (1-(-1)) /(3-0) = +2/-3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,-1) set x = 0, y= -1 in y = (2/3)x + b =>

-1 = (2/3) 0 + b => b = -1

Therefore y = (2/3)x + -1

Thank you,

MrB

5 0

3 years ago

Cheyenne collects movies. She has 435 movies that are either comedy, scary, or sci-fi. She has the same number of videos for eac

tensa zangetsu [6.8K]

145 movies of each genre.

435/3= 145

4 0

3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago

Given the function f(x) = x3 + 2x2 − 3x − 5, what is resulting function when f(x) is shifted to the right 1 unit?

cluponka [151]

When shifted to the right 1 unit it would be:
f(x - 1) = (x - 1)^3 + 2(x - 1)^2 - 3(x - 1) - 5
= (x^3 - 3x^2 + 3x - 1) + 2(x^2 - 2x + 1) - 3(x - 1) - 5 
= x^3 - 3x^2 + 2x^2 + 3x - 4x - 3x - 1 + 2 + 3 - 5 
= x^3 - x^2 - 4x - 1
I hope this helps

5 0

3 years ago

Read 2 more answers

7x-3/4=x what is the process to answer fasttttttt please

blondinia [14]

Answer:

x= 1/8 is the correct answer

7 0

3 years ago

Read 2 more answers