Help me Please this is due Tonight Ill give you the best answer if you do After a rotation of 180° about the origin, the coordinates of the vertices of the image of a triangle are A'(6, 3), B'(-2, 1), and C'(1, 7). What is the y-coordinate of the pre-image for vertex A?

After a rotation of 180° about the origin, the coordinates of the vertices of the image of a triangle are A'(6, 3), B'(-2, 1), and C'(1, 7). What is the x-coordinate of the pre-image for vertex A?

john got lost so he didn't know what to do and it was crazy because he didn't except it

step-by-step explanation:

123123 step byt step

6e14 is the answer.

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west