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:
90 km, N 46° E
Step-by-step explanation:
<em>A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.</em>
Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).
This will be the jet's distance from its starting point.
In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.
The <u>Law of Cosines</u> will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.
In this case, the known angle is 110° (angle B) so we will use the <u>Law of Cosines</u> respective to B.
Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).
- b² = (60)² + (50)² - 2(60)(50) cos(110°)
- b² = 6100 -(-2052.12086)
- b² = 8152.12086
- b = 90.28909602
- b ≈ 90 km
The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.
The <u>Law of Cosines</u> with respect to A:
Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).
- (60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
- 3600 = 8152.12086 - 6528.909602 cosA
- -4552.12086 = -6528.909602 cosA
- 0.6972252853 = cosA
- A = cos⁻¹(0.6972252853)
- A = 45.79519
- A ≈ 46°
The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.
All you have to do is multiply 2.3 by 13 together to get 29.9. Then you would subtract that by the total volume which would get 107.64. That's the width.
Answer is in the photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
tinyurl.com/wpazsebu
Answer:
i think 6x
Step-by-step explanation:
cz we have a variable
