2x + 10 and x + 40 are vertical angles. By definition, vertical angles are congruent. So let's take care of that first. 2x+10 = x+40. Subtract an x from both sides and at the same time subtract 10 from both sides to get x = 30. Ok so x = 30. So what, right? We need y. Well, 2y+x is vertical with an angle that has the exact same measure as 2y+x because vertical angles are congruent. We know that all those angles add up to equal 360, so here's our equation:

. Combining like terms gives us 5x+4y=360. We already have a value for x, remember, so we will sub it in. x=30. 5(30)+4y=360 and 150 + 4y = 360. Subtract 150 from both sides and we have 4y = 210. Divide both sides by 4 to get that y = 52.5. There you go!
http://www.stgeorges.co.uk/blog/how-can-i-describe-a-graph-ielts-writing-task-part-1-business-english
Answer:
(1, 1)
Step-by-step explanation:
Given vertices of the <u>parallelogram</u>:
- H = (-1, 4)
- J = (3, 3)
- K = (3, -2)
- L = (-1, -1)
Therefore the <u>parallel sides</u> are:

Therefore, the <u>diagonals</u> of the parallelogram are:

To find the <u>coordinates of the intersection of the diagonals</u>, either:
- draw a diagram (see attached) and determine the point of intersection of the diagonals from the diagram, or
- determine the midpoint of either diagonal (as the diagonals of a parallelogram bisect each other, i.e. divide into 2 equal parts).
<u>Midpoint between two points</u>

To find the <u>midpoint of diagonal LJ</u>, define the endpoints:
Substitute the defined endpoints into the formula and solve:

Therefore, the coordinates off the intersection of the diagonals of parallelogram HJKL are (1, 1).
Learn more about midpoints here:
brainly.com/question/27962681
A simple rule to tell whether a graph is a function is to pretend to cut it vertically, if any cut intercept the graph twice, it is not a function. otherwise, it is.
In this case, the heart is not a function, because if you cut it up-down direction, your imaginary knife will touch the edge of the heart twice.
I hope this explanation helps.<span />