Answer:
6T + p
6T + p = 27
T = 4 ; p = 4
Step-by-step explanation:
Given the following :
Points awarded for touchdown = 6 and then try for a point 'p' after touchdown
Hence, number of points scored on
touchdowns T and points after touchdowns p by one team in a game.
6points * number of touch downs (T) + point 'p' after touchdown
= 6T + p
B) If a team wins a football game 27-0,
Possible number of touch downs and points after touchdown :
Then 6T + p = 27
27 / 6 = 4 remainder 3
Hence, possible number of touch downs = 4
Number of points p = 27 - (6*4) = 27 - 24 = 3
C.) if winning = 21-7
Team A = 21
Hence, T = 21/6 = 3 remainder 3
T = 3 ; p = 3
Team B = 7
T = 7/6 = 1 remainder 3
T = 1 ; p = 1
Hence,
Total for both teams :
T = (1 + 3) = 4
p = (3 + 1) = 4
Answer:
2x + 8
Step-by-step explanation:
solve by using the distributive property
2(x + 4) = (2)(x) + (2)(4)
2x + 8
j^3
_______
4^2 h^4
when you have negatives, you want them to be positive. To do that you need to convert it into a fraction
