Easy
y=a(x-h)^2+k
vertex is (h,k)
we know that vertex is (4,0)
input that point for (h,k)
y=a(x-4)^2+0
y=a(x-4)^2
passes thorugh the point (6,1)
input that point to find a
1=a(6-4)^2
1=a(2)^2
1=a(4)
divide both sides by 4
1/4=a
thefor the equation is
y=(1/4)(x-4)^2
or
y=(1/4)x^2-2x+4
<h3>Given</h3>
trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8
<h3>Find</h3>
the area of PSTK
<h3>Solution</h3>
It helps to draw a diagram.
∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as
... KS² = PS² + KP²
... 13² = PS² + 12²
... PS = √(169 -144) = 5
This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is
... A = (1/2)(b1 +b2)h
... A = (1/2)(12 +8)·5
... A = 50
The area of trapezoid PSTK is 50 square units.
Part 1:
Given that the length of the chord is 18 cm and the chord is midway the radius of the circle.
Thus, half the angle formed by the chord at the centre of the circle is given by:

Now,

Therefore, the radius of the circle is
10.4 cm to 1 d.p.
Part 2I:
Given that the radius of the circle is 10 cm and the length of chord AB is 8 cm. Thus, half the length of the chord is 4cm. Let the distance of the mid-point O to /AB/ be x and half the angle formed by the chord at the centre of the circle be θ, then

Now,

Part 2II:
Given that the radius of the circle is 10cm and the angle distended is 80 degrees. Let half the length of chord CD be y, then:

Thus, the length of chord CD = 2(6.428) = 12.856 which is approximately
12.9 cm.
Answer:
5.6
Step-by-step explanation:
( 9 + 6 + 3 + 9 + 7 + 2 + 1 + 5 + 6 + 8 ) / 10
= 56 / 10
= 5.6