Write an equation for a line that is perpendicular to the line 6y – 9x = 12

Step-by-step explanation:

You need one more piece of information to get a specific line. I'll get the general one and then we'll see what comes of it.

Add 9x to both sides.

6y - 9x + 9x = 9x + 12

6y = 9x + 12                   Divide by 6

6y/6 = 9x/6 + 12/6

y = 1.5x + 2

Now the perpendicular line must have a slope that is found by using

m1 * m2 = - 1

m1 = 1.5             From the equation above.

1.5 * m2 = - 1     Divide by 1.5

m2 = -1/1.5

m2 = - 2/3

So far what you have is

y = -2/3 x + b                  You need a value for b. Usually that is found with a point. But I will just call b = 1 since there is nothing else to do

The complete equation becomes

y = -2/3 x + 1

The graph is enclosed.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.