Hold on, does the equation look like this? (x^2+6)/x-6? Because if so then you'd substitute nine for x first. Then you'd multiply nine by itself getting eighty-one, after that add six to get eighty-seven. substitute nine for x one more and subtract six. take eighty-seven divided by three to get an answer of twenty-nine. But if the equation is saying the square of nine aside from nine squared then you'd have an entirely different problem. That one would look similar to *square symbol*(9)+6/9-6
, You'd solve this by finding the square of nine which is three then you'd add six and get a numerator of nine. Then you go about the bottom the same as you would have in the first equation, subtract six from nine giving you a denominator of three so you now have nine over three which simplified is three.
The two answers are 29 or 3 depending on the equation.
The answer to this problem is B. 1x105
Answer:
1 and 2 are not polyhedrons, as a circle has infinite sides
1. cone, 1 vertex
2. sphere, nothing
3. pentagonal prism?
wait but can't every face be a base???
anyway 7 faces, 10 vertices
4. triangular prism
5 faces, 6 vertices
Step-by-step explanation:
