<span>To do this, we have to setup equations and then use substitution to get like terms.
c=children, s=students, a=adults
For total cost: 5350 = 5c + 7s + 9a
Total capacity: 750 = c + s + a
Data given: (.5)(c + s) = a
Now, rearranging the data given equation for (c+s) gives us (c+s)= 2a. Using this and substituting into the total capacity equation for (c+s) will allow us to find a: 750 = 2a + a = 3a which gives us a = 750/3 = 250.
Next, we set a = 250 in total cost and data given equation to get the following:
Total Cost: 5350 = 5c + 7s + 2250 which gives 3100 = 5c + 7s
Total Capacity: 750 = c + s + 250 which gives 500 = c + s
Rearranging total capacity to solve for s: s = 500 - c
Plugging that expression for s into total cost gives us: 3100 = 5c + 7(500-c) = 5c - 7c + 3500.
Solving for c gives us 400 = 2c . Total number of children is c = 200.
Total number of students is s = 750 - 200 - 250 = 300.</span>
31-62=152+2-6(3+21)
31-62=152+2-18+126
-31 =262
Answer:
Step-by-step explanation:
según el número de animales
Answer: the maximum is 25.
Step-by-step explanation: a max/min can occur on the endpoints of a function and critical points of the function's derivative.
f(x)=x^4-x^2+13
f'(x)=4x^3-2x
The critical points of f'(x) occur when f'(x) is zero or undefined. f'(x) is not ever undefined in this case, so we just need to find the x values for when it's zero.
0=4x^3-2x
x=.707, -.707
Now that we have the critical points of f'(x) (.707 and -.707) and endpoints (-1 and 2), we can plug in these x values into the original function to determine its maximum. When you do this you'll find that the greatest y value produced occurs when x=2 and results in a max of 25.
