<u>Question 4</u>
1) bisects , , and (given)
2) (an angle bisector splits an angle into two congruent parts)
3) and are right angles (perpendicular lines form right angles)
4) and are right triangles (a triangle with a right angle is a right triangle)
5) (reflexive property)
6) (HA)
<u>Question 5</u>
1) and are right angles, , is the midpoint of (given)
2) and are right triangles (a triangle with a right angle is a right triangle)
3) (a midpoint splits a segment into two congruent parts)
4) (HA)
5) (CPCTC)
<u>Question 6</u>
1) and are right angles, bisects (given)
2) (reflexive property)
3) (an angle bisector splits an angle into two congruent parts)
5) (HA)
6) (CPCTC)
7) bisects (if a segment splits an angle into two congruent parts, it is an angle bisector)
<u>Question 7</u>
1) and are right angles, (given)
2) and are right triangles (definition of a right triangle)
3) (vertical angles are congruent)
4) (transitive property of congruence)
6) (HA theorem)
7) (CPCTC)
8) bisects (definition of bisector of an angle)
Answer:
A
Step-by-step explanation:
Equation for a circle of radius r, centered at (h,k):
(x-h)² + (y-k)² = r²
£1 = 25 hits
Will get the equation is
so, X=25
x = What we want to find
2x + 3y = 12
Standard form is Ax + By = C, A, B and C must be intergers, cannot be fractions or decimals
y = -2/3x + 4 First we want to get rid of the fractions. There is only one, so if we multiply both sides by 3 that should do it.
3y = -2x + 12 Now make the constant be on one side by itself