Equation of a parabola with vertex at (2, -1) is
y = a(x - 2)^2 - 1
Using the given point: -3 = a(4 - 2)^2 - 1
-2 = a(2)^2
4a = -2
a = -1/2
Therefore, required equation is
y = -1/2(x - 2)^2 - 1
y = -1/2(x^2 - 4x + 4) - 1
y = -1/2x^2 + 2x - 2 - 1
y = -1/2x^2 + 2x - 3
A regular trapezoid is shown in the picture attached.
We know that:
DC = minor base = 4
AB = major base = 7
AD = BC = lateral sides or legs = 5
Since the two legs have the same length, the trapezoid is isosceles and we can calculate AH by the formula:
AH = (AB - DC) ÷ 2
= (7 - 5) ÷ 2
= 2 ÷ 2
= 1
Now, we can apply the Pythagorean theorem in order to calculate DH:
DH = √(AD² - AH²)
= √(5² - 1²)
= √(25 - 1)
= √24
= 2√6
Last, we have all the information needed in order to calculate the area by the formula:

A = (7 + 5) × 2√6 ÷ 2
= 12√6
The area of the regular trapezoid is
12√6 square units.
I think it may be 5+3 and carry the one up top to get the sum of 56.
Answer: for a the answer is 140° and for b x=25°
Step-by-step explanation:
for a, when you have 2 parallel lines cut by a transversal,the corresponding angles are congruent. and the angle140 is congruent to the corresponding angle as they are vertical angles are congruent.
for b, angle x is congruent to the angle that is supplementary to 155 therefore 180-155=25°