(b+h)(b+h)= b2+2hb+h2.
b/h
You cannot. If the base is B and the height is H any of the following will give an area of 13: B = 13, H = 1 B = 130, H = 0.1 B = 1300, H = 0.01 etc. Or, B = 6.5, H = 2 B = 65, H = 0.2 etc Or B = 2.6, H = 5 B = 26, H = 0.5 etc I hope you get the idea.
v=B*H B= area of the Base so... v= (b*h)*h
b = Ah divide both sides by A b/A = h
213 Bones in the Human Body
(b+h)(b+h)= b2+2hb+h2.
b/h
There is no figure given!!!! However, In a triangle the three angles are A B & 90 degrees. The sides opposite to A,B, & 90 are 'a' , 'b' & 'h' respectively. Hence Sin A = a/h SinB = b/h CosA = a/h ( Check ; 'a/h') Csc B ( CosecantB) = 1/ (a/h) = h/a CotB = CosB / SinB = (a/h) / (b/h) = a/b
You cannot. If the base is B and the height is H any of the following will give an area of 13: B = 13, H = 1 B = 130, H = 0.1 B = 1300, H = 0.01 etc. Or, B = 6.5, H = 2 B = 65, H = 0.2 etc Or B = 2.6, H = 5 B = 26, H = 0.5 etc I hope you get the idea.
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A = h/2*(a + b) So 2A/h = a + b and therefore, a = 2A/h - b
v=B*H B= area of the Base so... v= (b*h)*h
Area of a parallelogram = b*h. Call the area of parallelogram one x and two y. We now know that x = y = b(x)h(X) = b(x)h(y). Now b(x)h(x) = b(x)h(y) b(x)h(x)/b(x) = b(x)h(y)/b(x) h(x) = h(y), so the height of the second parallelogram must equal the height of the first.
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