Theorem 5
Let f be a differentiable function on I and let x 0 be any interior point of I. Then (a) If f attains its absolute maximum value at x 0 , then f ' (x 0 )= 0 (b) If f attains its absolute minimum value at x 0 , then f '(x 0 ) = 0. In view of the above theorems, we state the following rul..
Theorem5
The sum of the three angles of a triangle is 180 o..
Theorem 5
A rhombus is a parallelogram whose diagonals meet at right angles. ABCD is a rhombus. AC and BD meet at right angle..
A rhombus is a parallelogram whose diagonals meet at right angles. ABCD is a rhombus. AC and BD meet at right angle..Rectilinear Figures Theorem 5
Theorem 5 - A rhombus is a parallelogram whose diagonals meet at right angles. ABCD is a rhombus. AC and BD meet at right angle..
Theorem 5 - A rhombus is a parallelogram whose diagonals meet at right angles. ABCD is a rhombus. AC and BD meet at right angle..Converse of Theorem5
If the diagonals of a parallelogram are perpendicular then it is a rhombu..
Theorem 5(a):
If B is a Boolean algebra, for any x, y in B, then 0' = ..
Lami's Theorem
Lami's Theorem - If three concurrent forces acting on a body keep it in equilibrium, then each force is proportional to the sine of angle between the other two forces. Þ If three concurrent forces keep a body A in equilibrium, the 3 forces must be coplana..
Lami's Theorem - If three concurrent forces acting on a body keep it in equilibrium, then each force is proportional to the sine of angle between the other two forces. Þ If three concurrent forces keep a body A in equilibrium, the 3 forces must be coplana..Theorem 1
Theorem 1 - Parallelograms on the same base and between the same parallels are equal in area. ABCD and ABEF are two parallelograms on the same base AB and between the same parallels AB, DE area (ABCD) = area (ABEF) An alternate way of proving this theorem is as follows: Since bo..
Theorem 1 - Parallelograms on the same base and between the same parallels are equal in area. ABCD and ABEF are two parallelograms on the same base AB and between the same parallels AB, DE area (ABCD) = area (ABEF) An alternate way of proving this theorem is as follows: Since bo.. Result
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