Answer:
(I am assuming this is just approximation since there are no other numbers, just to be clear :))
Algebra: about 450 students
Geometry: about 250 students
Calculus: about 50 students
So, one thing that we would really want to do is to take (first), the whole numbers. This make it a whole lot easier for the both of us. So we do
. . .
Then we add up the fractions.
Your answer: (4)
Answer:
-11, -10, -9
Step-by-step explanation:
The middle one is their average, -30/3 = -10. The three integers are -11, -10, -9.
_____
I like to work "consecutive integer" problems using the average value of the integers involved. The average of <u>two</u> consecutive integers is 1/2 more than the small one, and 1/2 less than the large one, for example. Here, there are <u>three</u> consecutive integers, so their average is the middle one.
If you want to write that using an equation, you can let x represent the middle integer. Then the sum is ...
(x -1) +(x) + (x+1) = 3x = -30
x = -30/3 = -10 . . . . . the same average we computed above.
x -1 = -11 . . . . . the smaller of the other two integers
x +1 = -9 . . . . . the larger of the other two integers
Answer:
y=4x-5
Step-by-step explanation:
-3y = -12x+15. y = (-12x+15)/(-3). y = (-12x/-3)+(15/-3). y = 4x-5.