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Maslowich
3 years ago
7

WILL GIVE BRAINLIST AS SOON AS I CAN... is this A? i think it is but i might be wrong.​

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

I think 'neither' because A and C do not connect anywhere, while they are NOT parellel.

Hope this helps.

igor_vitrenko [27]3 years ago
5 0

Answer:

The answer is B, they are perpendicular

Step-by-step explanation:

To be able to figure this out, you have to first find the relationship between a and b.

You can conclude lines a and b are parallel by the consecutive angles theorem. Line d would be the transversal line that intercepts line a and line b. Because the consecutive angles created are equal, lines a and b are parallel .

Then you can conclude the relationship between line c and d through the exact same principle. Lines c and d are parallel.

Notice how line a and line d create a right angle (as indicated by the square drawn on the angle) which means the lines are perpendicular to each other.

Since line c is parallel to line d, you can finally conclude that the intercept of line c and a would create a right angle and the lines would this be perpendicular.

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For every 3 girls in Mr Hegarty's class there are 2 boys. What is the ratio of girls to boys in the class? Give your answer in i
kirill115 [55]

Answer:2boys for five studenrs so 2/5 and for girls 3/5 so response is 3/5


Step-by-step explanation:


7 0
3 years ago
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Use cross products to find the area of the triangle in the xy-plane defined by (1, 2), (3, 4), and (−7, 7).
Free_Kalibri [48]

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 a \times d 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.


S = ad - \frac 1 2 ab -  \frac 1 2 cd - \frac 1 2 (a-c)(d-b) = \frac 1 2(2 ad - ab -cd - ad +ab +cd -bc) = \frac 1 2(ad -bc)


That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) will have the same area as one whose vertex C is translated to the origin.


We set a=x_1 - x_3, b= y_1  - y_3, c=x_2 - x_3, d=y_2- y_3


S= ad-bc=(x_1 - x_3)(y_2 - y_3) -(x_2-x_3)(y_1 - y_3)


That's a perfectly useful formula right there. But it's usually multiplied out:


S= x_1y_2 - x_1 y_3  - x_3y_2 + x_3 y_3 - x_2 y_1 + x_2y_3 + x_3 y_1 - x_3 y_3


S= x_1 y_2 - x_2 y_1  + x_2y_3 - x_3y_2   + x_3 y_1 - x_1 y_3


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)

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3 years ago
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sergejj [24]

Answer:

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