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L=sqr((1/2 a+b+c) * (s-a) * (s-b) * (s-c))
If you mean a circle inscribed in the square: C = circumference, π = pi, r = radius, s = side, P = perimeter C = 2πr r = s/2 C = πs s = P/4 C = πP/4 So for this problem, the circumference is 9π, or about 28.3 If you mean a square inscribed in the circle, the computations are practically the same, except: r = sqrt(2)s/2 C = sqrt(2)πs C = sqrt(2)πP/4 So for this problem, the circumference is sqrt(2)9π, or about 40.0
If the bases have sides of length a, b and c units and the length of the prism is d units thenlateral area = (a+b+c)*darea of base = sqrt{s*(s-a)*(s-b*(s-c)} where s = (a+b+c)/2Then total surface area = lateral area + 2*area of base.
To find the area of a triangle with side lengths 12 cm, 7 cm, and 10 cm, we can use Heron's formula. First, calculate the semi-perimeter of the triangle by adding the three side lengths and dividing by 2: (12 + 7 + 10) / 2 = 29/2 = 14.5 cm. Then, use Heron's formula: Area = √[s(s-a)(s-b)(s-c)], where s is the semi-perimeter and a, b, and c are the side lengths. Plugging in the values, we get Area = √[14.5(14.5-12)(14.5-7)(14.5-10)] = √[14.52.57.5*4.5] = √2446.25 ≈ 49.46 cm².
S 2/x d/x bring the constant 2 out in front of the sign of integration 2 S 1/x dx you should know the integration of 1/x 2*ln(x) + C