Answer:
Solution given:
f(x)=x²
g(x)=x+5
h(x)=4x-6
now
23:
(fog)(x)=f(g(x))=f(x+5)=(x+5)²=x²+10x+25
24:
(gof)(x)=g(f(x))=g(x²)=x²+5
25:
(foh)(x)=f(h(x))=f(4x-6)=(4x-6)²=16x²-48x+36
26:
(hof)(x)=h(f(x))=h(x²)=4x²-6
27;
(goh)(x)=g(h(x))=g(4x-6)=4x-6+5=4x-1
28:
(hog)(x)=h(g(x))=h(x+5)=4(x+5)-6=4x+20-6=4x-14
Answer:
Hewo!
Step-by-step explanation:
im not sure of the answer but i got the trick!
so for the length add 8 to the width and for the area multiply the width by 14
This should get you the correct answers :)
Hope i helped! :D
Answer:
Option 3 - ![y=-6x+28](https://tex.z-dn.net/?f=y%3D-6x%2B28)
Step-by-step explanation:
Given : Perpendicular to the line
; containing the point (4,4).
To Find : An equation for the line with the given properties ?
Solution :
We know that,
When two lines are perpendicular then slope of one equation is negative reciprocal of another equation.
Slope of the equation ![x - 6y = 8](https://tex.z-dn.net/?f=x%20-%206y%20%3D%208)
Converting into slope form
,
Where m is the slope.
![y=\frac{x-8}{6}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bx-8%7D%7B6%7D)
![y=\frac{x}{6}-\frac{8}{6}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bx%7D%7B6%7D-%5Cfrac%7B8%7D%7B6%7D)
The slope of the equation is ![m=\frac{1}{6}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B1%7D%7B6%7D)
The slope of the perpendicular equation is ![m_1=-\frac{1}{m}](https://tex.z-dn.net/?f=m_1%3D-%5Cfrac%7B1%7D%7Bm%7D)
The required slope is ![m_1=-\frac{1}{\frac{1}{6}}](https://tex.z-dn.net/?f=m_1%3D-%5Cfrac%7B1%7D%7B%5Cfrac%7B1%7D%7B6%7D%7D)
![m_1=-6](https://tex.z-dn.net/?f=m_1%3D-6)
The required equation is ![y=-6x+c](https://tex.z-dn.net/?f=y%3D-6x%2Bc)
Substitute point (x,y)=(4,4)
![4=-6(4)+c](https://tex.z-dn.net/?f=4%3D-6%284%29%2Bc)
![4=-24+c](https://tex.z-dn.net/?f=4%3D-24%2Bc)
![c=28](https://tex.z-dn.net/?f=c%3D28)
Substitute back in equation,
![y=-6x+28](https://tex.z-dn.net/?f=y%3D-6x%2B28)
Therefore, The required equation for the line is ![y=-6x+28](https://tex.z-dn.net/?f=y%3D-6x%2B28)
So, Option 3 is correct.
Should have the ordered pairs of
(-4,4) (-3,3) (-2,2) (-1,1) (0,0) (1,1) (2,2) (3,3).
Will look like a V when graphed. Reason being is if you take the absolute value of anything you are basically just dropping the negative sign as it is how far away it is from 0.
Answer:
The second line (the flat one with an angle)
Step-by-step explanation:
Functions are linear when put on a graph.