In a dance competition, a participant has to score a total of at least 30 points in the first four rounds combined to move on to the fifth and final round. steward scored 5 points in the first round. he then went on to score additional points in the second, third, and fourth rounds. in each of those rounds, his score was identical. which inequality best shows the number of points, p, that steward scored in each of the second, third, and fourth rounds if he earned a place in the finals?

p ≥ 8.33

Step-by-step explanation:

Steward scored 5 points in the first round, then he needed 30 - 5 = 25 points in the following 3 rounds. Let's call p the three identical scores of these rounds. Therefore, the combined score of these rounds were 3*p, and must be equal-to-or-greater-than 25 to gain a place in the finals, that is:

3*p ≥ 25

p ≥ 25/3

p ≥ 8.33

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 ☺

mia khalifa

step-by-step explanation: