The answer is y=3x.
y=21, x=7
y=mx (m is the variable for slope)
21=m(7)
m=21/7
m=3
Hopefully this helps if not use Photomath
Answer:
x = students
y = adults
Linear Systems of Equations :
x + y = 937
0.75x + 2y = 1,109
Step-by-step explanation:
Step 1 : eliminate one of the variables
x + y = 937
0.75x + 2y = 1,109
-2(x + y = 937) = -2x - 2y = -1874
-2x - 2y = -1874
0.75x + 2y = 1109
We can eliminate y since -2y + 2y = 0
Step 2: Find x number of students.
-2x = -1874
+0.75x = 1109
____________
-1.25x = -765
____________ = 612 students attended
-1.25
937 - 612 = 325 adults
Check Answer :
612 x 0.75 = $459
325 x 2 = $650
$650 + $459 = $1,109
Therefore, there will be 612 students attending and 325 adults.
The questions for this problem would be:
1. What is the dimensions of the box that has the maximum volume?
2. What is the maximum volume of the box?
Volume of a rectangular box = length x width x height
From the problem statement,
length = 12 - 2x
width = 9 - 2x
height = x
where x is the height of the box or the side of the equal squares from each corner and turning up the sides
V = (12-2x) (9-2x) (x)
V = (12 - 2x) (9x - 2x^2)
V = 108x - 24x^2 -18x^2 + 4x^3
V = 4x^3 - 42x^2 + 108x
To maximize the volume, we differentiate the expression of the volume and equate it to zero.
V = 4x^3 - 42x^2 + 108x
dV/dx = 12x^2 - 84x + 108
12x^2 - 84x + 108 = 0x^2 - 7x + 9 = 0
Solving for x,
x1 = 5.30 ; Volume = -11.872 (cannot be negative)
x2 = 1.70 ; Volume = 81.872
So, the answers are as follows:
1. What is the dimensions of the box that has the maximum volume?
length = 12 - 2x = 8.60
width = 9 - 2x = 5.60
height = x = 1.70
2. What is the maximum volume of the box?
Volume = 81.872
