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:
Step-by-step explanation:
(8x²-18x+10)/(x²+5)(x-3)
express the expression as a partial fraction:
(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +bx+c/x²+5
both denominator are equal , so require only work with the nominator
(8x²-18x+10)=(x²+5)A+(x-3)(bx+c)
8x²-18x+10= x²A+5A+bx²+cx-3bx-3c
combine like terms:
x²(A+b)+x(-3b+c)+5A-3c
(8x²-18x+10)
looking at the equation
A+b=8
-3b+c=-18
5A-3c=10
solve for A,b and c (system of equation)
A=2 , B=6, and C=0
substitute in the value of A, b and c
(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +(bx+c)/x²+5
(8x²-18x+10)/[(x^2+5)(x-3)] = 2/x-3 + (6x+0)/(x²+5)
(8x²-18x+10)/[(x^2+5)(x-3)] =
<h2>2/(x-3)+6x/x²+5</h2>
(4x+2)/[(x²+4)(x-2)]
(4x+2)/[(x²+4)(x-2)]= A/(x-2) + bx+c/(x²-2)
(4x+2)=a(x²-2)+(bx+c)(x-2)
follow the same step in the previous answer:
the answer is :
<h2>(4x+2)/[(x²+4)(x-2)]= 5/4/(x-2) + (3/2 -5x/4)/(x²+4)</h2>
Answer:
Yes.
Step-by-step explanation:
Reduce both fractions.
Answer:
40%
Step-by-step explanation:
divide the students by the total
22/55
=
P
P
=
0.4 = 40%
Answer:
See the attached
Step-by-step explanation:
The solution and filled in Venn diagram is attached