Which of the following best forms the figure shown? 2 line segments that meet at an endpoint » 2 lines that meet at an endpoint 2 rays that meet at an endpoint 2 angles that meet at an endpoint

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

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.