4s+7a=861
s+a=168
This can be solved using either elimination or substitution. I am going to use substitution.
Solve s+a=168 for s
s=168-a
Replace 168-a for s in 4s+7a=861
4(168-a)+7a=861
672-4a+7a=861
Solve for a
672+3a=861
3a=189
a=63
Substitute 63 for a in s=168-a
s=168-63=105
So, s=105 student tickets and a=63 adult tickets
2x-y=4
Therefore: y=2x-4
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5x-10/4(2x-4)=10
5x-5/2(2x-4)=10
5x-5x+10=10
10=10
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*Problem can't be solved.
Answer:
Total cost of Mark's board is $3.75.
Step-by-step explanation:
We are given the following in the question:
Length of wooden board =
Unit cost of board = $0.50 per foot
Total cost of board =
Thus, total cost of Mark's board is $3.75.
Answer:
3m + 4c
Step-by-step explanation:
Whenever a word problem says the word earn that means the slope, also known as the rate of change, will be positive. Knowing this you can determine that both the caramel and milk chocolate slopes will be positive. After figuring all that out the only thing left to do is to make the equation. You know you have two slopes, and each slope needs a variable, so you will have to look back at the question. It is given that m represents the milk chocolate and c represents the caramel. Now all you have to do is make the slope the coefficient to the corresponding variable. The milk chocolates are 3 dollars, so the 3 goes in front of the m and the caramel chocolates are 4 dollars, so teh 4 goes in front of the 4. Since both slopes are positive no negatives or minus signs will be used in the equation. Knowing all this information you can now create the expression 3m + 4c.
Answer:
Step-by-step explanation:
Suppose the dimensions of the playground are x and y.
The total amount of the fence used is given and it is 780 ft. In terms of x and y this would be 3x+2y=780 (we add 3x because we want it to be cut in the middle). Therefore, y= 780/2-3/2x. Now, the total area (A )to be fenced is
A=x*y= x*(390-3/2x)=-3/2 x^2+390x
Calculating the derivative of A and setting it equals to 0 to find the maximum
A'= -3x+390=0
This yields x=130.
Therefore y=780/2-3/2*130=195
Thus, the maximum area is 130*195=25,350ft^2
