Answer:
diagonal of a polygon is a line from a vertex to a non-adjacent vertex. So a triangle, the simplest polygon, has no diagonals. You cannot draw a line from one interior angle to any other interior angle that is not also a side of the triangle.
Step-by-step explanation:
To find the length of the diagonal (or hypotenuse) of a right triangle, substitute the lengths of the two perpendicular sides into the formula a2 + b2 = c2, where a and b are the lengths of the perpendicular sides and c is the length of the hypotenuse. Then solve for c
Answer:
C. Multiplication
Step-by-Step Explanation:
This problem is testing your knowledge of the Order of Operations. Here is the correct order:
Exponents (inside parenthesis)- There are no exponents for you to solve in this expression, so just ignore this step for now. (example: (4 + 2²) --> (4 + 4))
Parenthesis (from left to right)- In this case, you would solve the expressions inside the parenthesis <em>first</em> because there are no exponents.
Multiplication and Division (from left to right outside of parenthesis)- After solving the expressions inside the parenthesis, there will be one multiplication you need to solve next.
Addition and Subtraction (from left to right outside of parenthesis)- The only operation left for you to do now is addition. You can either do this in one whole step, or you can divide it into two steps. I've done it in two to keep it simple.
Here is the problem worked out in steps in case you are confused:
Step 1: Parenthesis
(-8 ÷ 4) + 5 + 4(3 - 6)
(-2) + 5 + 4(-3)
Step 2: Multiplication
-2 + 5 + 4(-3)
-2 + 5 + (-12)
Step 3: Addition
-2 + 5 + (-12)
3 + (-12)
-9
(Note: '3 + (-12)' could also be written as '3 - 12', making it a subtraction instead of an addition)
Hope this helps!
Answer:
zotfnKhxitfupoydkfslfndckv
Answer:
Step-by-step explanation:
I'm assuming that strange looking thing is a 9, so begin by multiplying both sides by that denominator to get:
then subtract 3 from both sides to get
and then square both sides to get
and get everything on one side to solve for x by factoring:
and factor that however it is you have learned to factor second-degree polynomials to get that
x = 16 and x = 9. We have to check for extraneous solutions because anytime you manipulate an equation, as we did by squaring both sides, you run the risk of these solutions that actually don't work when you plug them back into the original equation. Let's try 16 first:
If 16 is a solution, then this statement will be mathematically true.
and 
so 16 works. Let's try 9:
We know that one doesn't work, because 9-9 = 0 and 0 over anything = 0, not 1.
x = 16 is the only solution.
