7) Area trapezoid = (B+b).H/2, but the Median is equal to (B+b)./ 2
84 =12 . H ==> H = 84/12 ==> H = 7
11) Area Equilateral triangle inscribe in a cercle with Radius R = 2√3
Area = (B.. Altitude) / 2. Calculate H, the altitude. The altitude in an equilateral triangle bisects the opposite side. Apply Pythagoras
(2√3)² = (√3)² + H² ==> 12 = 3 + H² ==> H² = 9 & H = 3
Hence Are = (2√3 x 3) /2 ==> 3√3 unit² or 5.2 unit²
12) Area of regular hexagone with perimeter =12
regular hexagone is formed with 6 equilateral triangles with
each side =12/6 = 2 units
Let's calculate the area of 1 equilateral triangle. Follow the same logic as in problem 11 & you will find that Altitude = √3, Area =(2.√3)/2 =√3 =1.73 Unit³
13) a) Circumference of a circle : 2πR==> 2π(30) =60π = 188.4
b) Area of a circle =πR³ ==> π(30)² = 900π = 2,826 unit²
1 to left gives (x -1)^2
vertical stretch makes it 5(x - 1)^2
reflection in axis changes the sign to - :-
equation of new function is f(x) = -5(x - 1)^2
The answer is n equals five
Answer:
800 ft
Step-by-step explanation:
2/3 minutes = 2/3 min * 60 sec/min = 40 sec = t
5 ft/sec = r
d= r * t = 5 ft/sec * 40 sec = 800 ft
