Recall the secant-tangent theorem, and you have
EA^2 = EC*CD
12^2 = 8*(x+10)
and now ED = EC+CD = 8+x+10
I suspect a typo somewhere in the murk above
If Hannah gives her younger sister 3 shirts, it does not matter what order she hands them to her. No matter the order, it will still be the same group of 3 shirt. Since order is not important this problem can be solved using a combination.
Specifically we are asked to find 8C3 (sometimes called "8 choose 3"). This is a fraction. In the numerator we start with 8 and count down 3 numbers. In the denominator we start with 3 and count all the way down to 1. Thus we obtain,
Answer:
It's fine. It is a function. The language can be very convoluted. Put simply a domain value cannot have 2 different range values.
Step-by-step explanation:
The confusing part I think, is probably the fact that -1 and 1 both have an answer of 5.
That's fine. Parabolas do that and they are functions. Two different xs can have the same value. The thing that eliminates functions is when the x value (the domain) has two different y values (the range values). Then you don't have a function.
To try and make it clearer if 1 pointed at both 3 and 5 you would not have a function.
The most accurate "estimation" for 5+4 is um... 9