13 vehicles (at least) will be needed to transport the students. 18x3= 54+10= 64\5= 12.8. Round 12.8 to the nearest while number is 13
well logically i dont think you could simplify this problem since both of your variables arent like terms... you couldnt add them because of the squared 2c. im sorry i couldnt help you
maybe check out khan academy? goodluck!
Answer:
Let's define:
A = # of students in group A
B = # of students in group B
C = # of students in group C.
"The total number of students who could attend a field trip is represented by the variable t."
This can be written as:
A + B + C = t.
"The number of students in Group A is less than the number in Group B."
Here we have a strictly "less than", then this is written as:
A < B.
"Group A has 6 students more than 1/4 the total number of students"
A = t/4 + 6
"Group B has 3 less than the total number of students"
B = t - 3.
Then we have the equations:
A + B + C = t.
A = t/4 + 6
B = t - 3.
A < B.
We could replace the second and third equatio in the fourth one, to get:
t/4 + 6 < t - 3.
t/4 + 9 < t
9 < t - t/4
9 < t*(4/4 - 1/4)
9 < t*(3/4)
(4/3)*9 < t
12 < t.
Then we found an inequality that defines the minimum possible value of t,
All the numbers that you multiply to get 60
Answer:
(fg)(x) = x^2 -2x -8
Step-by-step explanation:
(fg)(x) = f(x)·g(x) = (x +2)(x -4)
= x(x -4) +2(x -4) = x^2 -4x +2x -8
= x^2 -2x -8 . . . . matches the 3rd choice