Answer:
you cut off some of the problem
Step-by-step explanation:
C i thinks, but im not even sure so
<u>Question 6</u>
1) 
, 
, O is the midpoint of 
, 
(given)
2) 
are right angles (perpendicular lines form right angles)
3) 
are right triangles (a triangle with a right angle is a right triangle)
4) 
(a midpoint splits a segment into two congruent parts)
5) 
(LL)
<u>Question 7</u>
1) 
are right angles), 
2) 
(reflexive property)
3) 
are right triangles (a triangle with a right angle is a right triangle)
4) 
(LL)
5) 
(CPCTC)
<u>Question 8</u>
1) 
, point D bisects 
(given)
2) 
are right angles (perpendicular lines form right angles)
3) are right triangles (a triangle with a right angle is a right triangle)
4) 
(definition of a bisector)
5) (reflexive property)
6) (LL)
7) 
(CPCTC)
Answer:
53 + 3g ≤ 65, g ≤ 4
Step-by-step explanation:
To solve for g:
53 + 3g ≤ 65
First, subtract 53 from both sides.
53 - 53 + 3g ≤ 65 - 53
=> 3g ≤ 12
Divide 3 by both sides
=> 3g / 3 ≤ 12 /3
=> g ≤ 4
Therefore, g ≤ 4
Hope this helps :)
Answer:
True, false, true, true.
Step-by-step explanation:
The roots zeros of a quadratic function are the same as the factors of the quadratic function. This is true because your roots are your factors—>(x-3) is a factor, x=3 is the root.
The roots zeros are the spots where the quadratic function intersects with the y-axis. No! Those are called y-intercepts!
The roots zeros are the spots where the quadratic function intersects with the x-axis. True. X-intercepts are your solutions. (x-3) graphed would the (3,0). That’s a solution.
There are not always two roots/zeros of a quadratic function, True. No solution would be when your quadratic doesn’t intersect the x-axis. One solution would be when your vertex would be on the x-axis. Two solutions is when your quadratic intersects the x-axis twice. Can there be infinite solutions? No. It’s either 0, 1, or 2 solutions.
