PLEASE HELP The domain is {x∈R| x≠−3}, and the range is {y∈R| y≠−5}. The domain is {x∈R| x≠−2}, and the range is {y∈R| y≠−5}. The domain is {x∈R| x≠−5}, and the range is {y∈R| y≠−2}. The domain is all real numbers, and the range is all real numbers as well.

The domain is {x∈R| x≠−2}

The range is {y∈R| y≠−5}

Explanation:

Using the function, we can factor the numerator to get

x^2-x-6 = (x-3)(x+2)

Then the (x+2) terms cancel out leaving x-3 only

Therefore, f(x) simplifies to f(x) = x-3.

This produces the straight line graph that is shown. But there's a hole at (-2,-5)

Why is this? It's because plugging x = -2 into the original equation (before you simplify) leads to a division by zero error.

The denominator x+2 becomes -2+2 = 0 and we can't divide by zero.

If we plugged x = -2 into the simplified version of f(x), then we have

f(x) = x-3

f(-2) = -2-3

f(-2) = -5

Showing why the hole is at (-2,-5)

There is no way that x can equal -2, so this is why we kick it out of the domain. Similarly, there's no way to get to y = -5, and this value is kicked out of the range.

The notation x∈R means that x is in the real number set. Saying  {x∈R| x≠−2} means x is any real number but -2. The range is a similar story but we kick out -5 instead.

idk bring the function back   im not smart

step-by-step explanation:

(a) the volume of the cone is 20/3 in3

hence, option a is correct.