Obviously there is more than one way to do this. VL = Ldi/dt Volts has units of Joules/Coulomb: J/C i has units of Coulombs/second: C/s So di/di is C/s^2 L has units of J/C / C/s^2 = Js^2/C^2 Ic = CdV/dt => Ic/dV/dt = C/s / J/C-s = C/s * C-s/J = C^2/J C has units of C^2/J OR you could just type Q = CV => C = Q/V = C/J/C = C^2/J same answer R = V/I => J/C / C/s = J-s/C^2
It is sqrt{s*(s-a)*(s-b)*(s-c)} where the lengths of the three sides are a, b and c units and s = (a+b+c)/2.
Assuming 2, 8, and 8.5 are the lengths of the sides, the area is 7.929 cm^2. Using Heron's formula, A = sqrt(s * (s-a) * (s-b) * (s-c)) where a, b, and c are the lengths of the sides and s = (a + b + c)/2
The answer depends on what information you do have.Suppose you know only the lengths of the sides (a, b and c), then let s = (a + b + c)/2.Then area = sqrt[s*(s - a)*(s - b)*(s - c)]If 2 sides and the included angle, then area = 1/2*a*b*sin(C).There are other formulae.
The area is doubled. a,b - cathetus; c - hypotenuse; h - height; S - area. S = (a*b)/2 = (c*h)/2 obviously if k is the doubled height. and A is the new area. A = (c*k)/2 = (c*2h)/2 = c*h and A = S*2
2 Sides of a Coin
?What?
If the sides of the triangular base are a, b and c and the height is h, the area is 2sh + 2 sqrt (s(s-a)(s-b)(s-c)) where s = (a +b +c)/2 [I. e. s is the semiperimeter of the base] and sqrt means"the square root of "
Use the Hero's formula: Let s = (a + b + c)/2. Then the area of the triangle equals√[s(s - a)(s - b)(s - c)], where a, b, and c denote the sides of the triangle.
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
Suppose the sides are a, b and c units. Calculate s= (a+b+c)/2 Then Area = sqrt[s*(s-a)*(s-b)*(s-c)] square units
C. S. Forester died on April 2, 1966 at the age of 66.