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This is a Ditloid. The answer is 1 step forward, 2 steps back.
To calculate the area of a triangle you can use three different methods. 1) Area = 1/2 base x vertical height. 2) Area = √[s(s - a)(s - b)(s - c)]. Where s = 1/2(a + b + c) and a,b and c are the lengths of the sides of the triangle. 3) Area = 1/2ab sin C = 1/2ac sin B = 1/2bc sin A. Where a, b and c are the lengths of the sides of the triangle and A,B and C are the angles opposite the side with the same (lower case) name.
The B-S- of A- with Brian Sack - 2011 1-2 was released on: USA: 18 November 2011
For finding the area of triangle with the help of Heron's formula: [s(s-a)(s-b)(s-c)]1/2 where a,b,c are sides of the triangle and s=1/2 of perimeter of the triangle
No. The B-2 is a modern bomber. The B-1 was not completed until 1980's and the B-2 even later.
The next letter will be a "b." The pattern 2-1-3-2-4-3-etc. So the "a"s and "b"s will each increase by one. First there were two "b"s, then one "a." Then there were three "b"s and two "a"s. So next there will be four "b"s and three "a"s, five "b"s and four "a"s, and so on.
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.
L=sqr((1/2 a+b+c) * (s-a) * (s-b) * (s-c))
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
Here the sample space is(s)=10, =>mod(s)=10 a=be the event of getting exactly 5 boys =>mod(a)=5 b=be the event of getting exactly 5 girls =>mod(b)=5 thus, p(a)=mod(a)/mod(s)=5/10=1/2 p(b)=mod(b)/mod(s)=5/10=1/2 p(5 boys and 5 girls)=p(a)*p(b)=1/2*1/2=1/4
The two numbers are 18 and -24.If you're having trouble with the factoring, you can always use the quadratic formula:Let the two numbers be a & b. a*b = P and a+b = S {for Product and Sum}So substitute b = S-a {from the Sum formula}, and you have a*(S-a) = P, or:a*S - a² = P ----> a² - S*a + P = 0.So with the quadratic formula:a1 = (-S + sqrt(S^2 - 4*1*P)/(2*1) Anda2 = (-S - sqrt(S^2 - 4*1*P)/(2*1)Substituting -432 for P and -6 for S, we get a1 = 18, and a2 = -24. Note that substituting a1 into the original formulas, gives b = a2, or if you use a2, then you get b=a1. So the two numbers are a1 and a2.