Triangle PQR has vertices P(–3, –1), Q(–3, –3), and R(–6, –2). The triangle is rotated 90° counterclockwise using the origin as the center of rotation. Which graph shows the image, triangle P’Q’R’? On a coordinate plane, triangle P prime Q prime R prime has points (negative 3, negative 1), (negative 3, negative 3), (negative 6, negative 2). On a coordinate plane, triangle P prime Q prime R prime has points (negative 1, 3), (negative 3, 3), (negative 2, 6). On a coordinate plane, triangle P prime Q prime R prime has points (1, negative 3), (3, negative 3), (2, negative 6). On a coordinate plane, triangle P prime Q prime R prime has points (3, negative 1), (3, negative 3), (6, negative 2).

the answer is 3/5

step-by-step explanation:

first divide 24 divided by 8 because that is the least common multiple

then divide 40 by 8

then you will get 3/5.

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

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!

step-by-step explanation: