Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan
= 5/11.65
=23.23°
AC= 
km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
-1(7+4b) - distribute the -1 to the expression
(-1 * 7) + (-1 * 4b) — ( + and - = - )when it is multiplied
= -7 - 4b — there is no like term to added to subtracted so it will stay as like
x(x^2 - 2xy+y^2) distribute x to all
= x(x^2) - x(2xy) + x(y^2)
= x^3 - 2x^2y + xy^2 in multiplication exponent add to similar variable
No like term to connect
-5x(-3+x)
= -5x(-3) -5x(+x). (- and - is +)
= 15x -5x. similar variable x so connect the like connect like term by subtracting them
= 10x
I'm assuming that the diagonals intersect at E. Since diagonals of a parallelogram bisect each other, BE=ED.
- 7x-2=x²-10
- x²-7x-8=0
- (x-8)(x+1)=0
- x = -1, 8
As distance must be positive, we reject the negative case, so x=8.
Thus, BE=ED=54.
Answer:
Um the picture isn´t very clear it is hard to see it.
Step-by-step explanation:
at least on my end it is.
Original position:
A-(-8,-4)
B-(-6,3)
C-(-3,7)
D-(-2,-2)
Translation:
A'-(-4,-4)
B'-(-2,3)
C'-(1,7)
D'-(2,-2)
Vertex C will be in quadrant 1 (+,+) after being translated 4 unites to the right.
