Answer:
what do you need help wth ?
Step-by-step explanation:
Answer:
x = y = 22
Step-by-step explanation:
It would help to know your math course. Do you know any calculus? I'll assume not.
Equations
x + y = 44
Max = xy
Solution
y = 44 - x
Max = x (44 - x) Remove the brackets
Max = 44x - x^2 Use the distributive property to take out - 1 on the right.
Max = - (x^2 - 44x ) Complete the square inside the brackets.
Max = - (x^2 - 44x + (44/2)^2 ) + (44 / 2)^2 . You have to understand this step. What you have done is taken 1/2 the x term and squared it. You are trying to complete the square. You must compensate by adding that amount on the end of the equation. You add because of that minus sign outside the brackets. The number inside will be minus when the brackets are removed.
Max = -(x - 22)^2 + 484
The maximum occurs when x = 22. That's because - (x - 22) becomes 0.
If it is not zero it will be minus and that will subtract from 484
x + y = 44
xy = 484
When you solve this, you find that x = y = 22 If you need more detail, let me know.
Answer:
A. (f-g)(x) = -7x -3
Step-by-step explanation:
(f -g)(x) = f(x) -g(x) = (-3x -5) -(4x -2)
= -3x -5 -4x +2
= -3x -4x -5 +2
(f -g)(x) = -7x -3
Answer:
Dennis paid $82 and Connie paid $46.
Step-by-step explanation:
We can set up an equation by putting in variables, c representing how much Connie paid. Since we know that Dennis paid $36 more, we will also factor that in the equation.
c + c + 36 = 128
Where c + 36 represents the amount Dennis paid, and 128 represents the total amount paid as given in the question. We can start by adding like terms. 2c + 36 = 128
Now, we can subtract 36 from each side,
2c + 36 - 36 = 128 - 36
2c = 92
Divide each side by two,
2c/2 = 92/2
c = 46
Now, to make sure this is correct, let's substitute our c for 46 in our equation:
46 + 46 + 36 = 128
92 + 36 = 128
128 = 128
Therefore, our equation is correct, and Dennis paid $82 while Connie paid $46.
For any point to be in the first quadrant, it must have a positive "x" value and "y" value.
If x = 1 then y = 2, a point with both x and y positive values which would be in the First Quadrant.
