Answer:
see explanation
Step-by-step explanation:
A quadratic function in standard form is
y = ax² + bx + c (a ≠ 0 )
Given
y = - 3x² + 6x + 17 ← compare coefficients with standard form, then
a = - 3, b = 6, c = 17
Given the quadratic in standard form the the equation of the axis of symmetry is
x = - 
= - 
= 1
Equation of axis of symmetry is x = 1
Answer:
Step-by-step explanation:
Evaluate 6(5-(6-5)-3) = 6(5-1-3) {first solve innermost bracket}
= 6* (5-4) = 6*1 =6
<h3>Explanation:</h3>
1. PQ║TS, PQ ≅ TS, PT and QS are transversals to the parallel lines . . . given
2. ∠P ≅ ∠T . . . alternate interior angles at PT
3. ∠Q ≅ ∠S . . . alternate interior angles at QS
4. ΔPQR ≅ ΔTSR . . . ASA postulate
_____
You can use any pair of angles together with the sides PQ and TS. If you use the vertical angles and one of ∠T or ∠S, then you must invoke the AAS postulate for congruence, as the side is not between the two angles.
