I love these. It's often called the Shoelace Formula. It actually works for the area of any 2D polygon.
We can derive it by first imagining our triangle in the first quadrant, one vertex at the origin, one at (a,b), one at (c,d), with (0,0),(a,b),(c,d) in counterclockwise order.
Our triangle is inscribed in the 
rectangle. There are three right triangles in that rectangle that aren't part of our triangle. When we subtract the area of the right triangles from the area of the rectangle we're left with the area S of our triangle.
That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices 
will have the same area as one whose vertex C is translated to the origin.
We set 
That's a perfectly useful formula right there. But it's usually multiplied out:
That's the usual form, the sum of cross products. Let's line up our numbers to make it easier.
(1, 2), (3, 4), (−7, 7)
(−7, 7),(1, 2), (3, 4),
[tex]A = \frac 1 2 ( 1(7)-2(-7) + 3(2)-4(1) + -7(4) - (7)(3)
Answer:
m = v + 2
Step-by-step explanation:
m is Mags, and v is Vector
m (Mags' age) = v(Vector's age) + 2
Answer:
15 * 15 * 15 = 3375
Step-by-step explanation:
Volume of a cube = (side) * (side)* (side)
Answer:
The partial quotients are 100, 20 and 6. The quotient is 126.
Step-by-step explanation:
The dividend is 378 and the divisor is 3.
First find the factors of 3, near to 300.
The value 300 is less than 378, so record the partial quotient 100 and subtract 300 from 378. Repeat the same process until the dividend has been zero.
Now the remaining divide is 78.
First find the factors of 3, near to 78.
Since 90>78, therefore 30 is not a partial quotient. The value 60 is less than 78 , so record the partial quotient 20 and subtract 60 from 78.
Now the remaining divide is 18.
The number 6 is a partial quotient.
Therefore partial quotients are 100, 20 and 6. Add the partial quotient to find quotient.
