To find the inverse,
First, you would have to switch the domain and range. In other words, swap x and y:
h(x) = 3/ -x-2
y = 3/ -x-2
x = 3/ -y-2
Then, when you have this, simply solve for y:
x = 3/-y-2
x (-y-2) = 3/-y-2 (-y-2)
x(-y-2) /x= 3 /x
-y-2 = 3/x
-y = 3/x + 2
y = -3/x -2
So the inverse would be
h⁻¹(x)= -3/x -2
Answer:
6is the answer
Step-by-step explanation:
EFGH is twice the size of ABCD
<span><span><span>x = ± 2</span></span><span><span>Then the solution is </span><span>x = ± 2</span></span><span><span>hope this helps</span></span></span>
Answer:
I think you're suppose to add the angles
Step-by-step explanation:
correct me if I'm wrong because as you can see ab CD are connected together even if you're still stuck I think the best thing is to ask your teacher.
