1-9 are all numbers that are less than 10, and 1 is included because the number can be greater than or equal to 1.
Hope this helped!
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
Answer:D
Step-by-step explanation:
3r+ n^2 -r+ 5 -2n+ 2
= n^2+ (3r -r) -2n+ (5+ 2) (combine like terms)
= n^2+ 2r -2n+ 7
The final answer is n^2+ 2r -2n+ 7~
The correct answer is A. 2^3=8 and 2^6=64. 8/64=(1/8). This is equal to 2^(-3).
