Plans for a new apartment complex call for buildings directly across the fence from each other to be congruent. This computer printout shows Building A. If the vertices of Building B are located at (21,21) 2 x 1 , 2 y 1 , (22,22) 2 x 2 , 2 y 2 , (23,23) 2 x 3 , 2 y 3 , and (24,24) 2 x 4 , 2 y 4 , will Building B be congruent to Building A? Math item stem image CLEAR CHECK Yes, because Building B is a reflection of Building A. No, because the transformation applied was a dilation. Yes, because the same transformation was applied to all the vertices. No, because the translation will not be on the other side of the fence.

other answer is correct. i had used the radius of the radius somehow. a website:

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺

b). 4320 have a nice day