(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
h pencils have harder lead, and they don't smudge easily and your lines are light-colored. Your b pencils have soft, black lead, and they are used for blending and shading when drawing. The higher the number for your b pencils, the darker your lead and lines are going to be.
Use a B, 2B, or darker grade of pencil for your shading. H and 2h pencils will give you less depth.
(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
If an additional B is added to the reaction, the reaction will be more likely to proceed towards completion, as there will be more B molecules available to react with KL and H. This may lead to an increase in the amount of product formed, depending on the stoichiometry and conditions of the reaction.
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.
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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