The first one is wrong
(do you need help solving it?)
        
             
        
        
        
Answer:
Step-by-step explanation:
In order to determine the information you're being asked for, you need to complete the square on that quadratic. The first step is to move the constant over to the other side of the equals sign:

Here would be the step where, if the leading coefficient isn't a 1, you'd factor it out. But ours is a 1, so we're good there. Now take half the linear term (the term with the single x on it), square it, and add it to both sides. Our linear term is a -2.  Half of -2 is -1, and -1 squared is +1. We add +1 to both sides giving us this:

Now we'll clean it up a bit. The right side becomes a 4, and the left side is written as its perfect square binomial, which is the whole reason we did this. That binomial is
 (set equal to the 4 here). Now we'll move the 4 back over and set the whole thing back equal to y:
 (set equal to the 4 here). Now we'll move the 4 back over and set the whole thing back equal to y:

From this it's apparent what the vertex is: (1, -4),
the axis of symmetry is x = 1, and 
the y-intercept is found by setting the x's equal to 0 in the original  equation and solving for y. So the y-intercept is (0, -3).
Your choice for the correct answer is the very last one there.
 
        
             
        
        
        
From the 8th to the 13th there are five numbers, and the difference between the 8th and the 13th is 20-12.5=7.5, so the difference between two neighboring number is 7.5÷5=1.5
from the 13th and the 25th there are 12 numbers, so the total difference is 12*1.5=18
so the 25th term is 20+18=38
        
             
        
        
        
c is the answer i know this because im smart and because c is the answer 
 
        
             
        
        
        
Answer:
HL
Step-by-step explanation:
The hypotenuse and the "leg" are congruent...along with the 90 degree angle.
"The HL Postulate states that if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent." - Calcworkshop