The steps to construct a congruent angle to given angle jkl using a compass and straightedge are given below. arrange them in the correct order.

draw the points of intersection between each arc and the ray, or rays, it intersects.
2.open the compass to the width of the intersection points between the arc and the two rays of given angle jkl.
3.use a straightedge to create a ray with endpoint t.
4.place the tip of your compass on the vertex of the given angle, point k, and swing an arc that intersects both rays of the given angle.
5.keeping the compass the same width, use the compass to draw an arc from the endpoint t.
6.draw a point where the two arcs intersect.
7.keeping the compass width the same, draw an arc from the point of intersection between the second arc and ray.
8.use your straightedge to draw a ray between endpoint t and the point of intersection between the two arcs.

6, 7, 1, 2, 8, 5, 4, 3

3, 4, 5, 1, 2, 7, 6, 8

8, 4, 3, 6, 1, 5, 2, 7

4, 6, 7, 1, 2, 5, 8, 3

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.