The probability of ac and bc is 1/5.
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i
since BC=1/2 AC you take half of 54 then solve the equation. 3x^2=54/2 3x^2=27 x^2=9 x=3
All the trigonometric functions are derived from the right angled triangle. If we consider the three sides (AB, BC, CA) of a triangle and the included angle. There is a possibility of getting six functions based on the ratios like AB/AC, BC/AC, AB/BC, BC/AB, AC/BC, AC/AB . So we will have six trigonometric functions
There is no distributive property of addition over multiplication. The equation works if a + (b * c) = (a + b)*(a + c) = a2 + ab +ac +bc => a + bc = a2 + ab +ac +bc ie a = a2 + ab + ac = a*(a+b+c) and that, in turn requires that a = 0 or a+b+c = 1 If a, b and c are fractions than the second condition requires the fractions to sum to 1 - not be equal to 1.
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A+BC+AC+B=A+BC+AC+B unless any of these variables has an assigned value.
The probability of ac and bc is 1/5.
If these are sides of a triangle then AC can have any value in the interval (3, 13).
AC=5 AB=8 A=1 B=8 C=5 BC=40
Assuming that AB and AC are straight lines, the answer depends on the angle between AB and AC. Depending on that, BC can have any value in the range (22, 46).
yes because ab plus bc is ac
I believe it means Before
It is possible, depending on what on earth AC and BC are!
ac is After Christ and bc is Before Christ
Take a right angled triangle ABC with the right angle at B, so that AC is the hypotenuse. Let AC be 1 unit long. Using the angle CAB, the length of AC and the trigonometric ratios: sin = opposite/hypotenuse ⇒ sin CAB = AB/AC = AB/1 = AB cos = adjacent/hypotenuse ⇒ cos CAB = BC/AC = BC/1 = BC Using Pythagoras: AB2 + BC2 = AC2 ⇒ (sin CAB)2 + (cos CAB)2 = 12 ⇒ sin2θ + cos2θ = 1
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i