Step-by-step explanation:
Answer:
Here's what I've done:
Let ϵ>0ϵ>0 be given.
Now,
|f(x)−f(y)|=|tan−1x−tan−1y|=∣∣tan−1(x−y1+xy)∣∣
|f(x)−f(y)|=|tan−1x−tan−1y|=|tan−1(x−y1+xy)|
For non-negative x,y∈R,x,y∈R,
|f(x)−f(y)|=|tan−1x−tan−1y|=∣∣tan−1(x−y1+xy)∣∣≤|tan−1(x−y)|≤|x−y|<δ
|f(x)−f(y)|=|tan−1x−tan−1y|=|tan−1(x−y1+xy)|≤|tan−1(x−y)|≤|x−y|<δ
We can choose ϵ=δϵ=δ. Now, for negative x,y∈Rx,y∈R. Please, how do I go about it?
Answer:
-38
Step-by-step explanation:
Just looking at the coordinates given you can do process of elimination. A reflection along the y axis would multiply the x values by negative 1. This does not work
Choice b a translation 2 units left and 4 units up. So subtract 2 from x and add 4 to y. This also does not work.
Choice c is reflection along x axis which multiplies y values by negative 1. After looking this does not work.
Choice d 2 units right(add 2 to x) and 4 units down(subtract 4 from y) this method works for all points. So this is correct
It’s A = 1/2(b1+b2)h
Is this is what you want