Answer:
0(2/3; 0)
Step-by-step explanation:
Answer:
A (4-g)1.5
Step-by-step explanation:
hope this helps if you need a explanation just ask in the comments and I will.
Answer:
4x-8
Step-by-step explanation:
y=-x2+6x-8
y=4x-8
Answer:
PART A: Inequality (a)
Solve for y
The graph of y ≥ ⅓(8-x) is represented by the upper red line and all points in the shaded area below it. The line is solid because points on the line satisfy the conditions.
Inequality (b)
Solve for y
The graph of y ≥ 2 - x is represented by the lower blue line and all points in the shaded area above it. The line is solid because points on the line satisfy the conditions. The solution lies in the purple area. It consists of all combinations of x and y that make y ≥ ⅓(8 - x) and y ≥ 2 - x. A practical but not a mathematical condition is that all values of x and y must be zero or positive numbers (for example, you can't have -2 servings of food), so I have plotted only the numbers in the first quadrant.
PART B: If a point is a solution of the system, then the point must satisfy both inequalities of the system.
For x=8, y=2. Verify inequality A is not true. So the point does not satisfy inequality A. Therefore, the point is not included in the solution area for the system.
PART C: I choose the point (3,1) which is included in the solution area for the system.
That means Michelle buys 3 serves of dry food and 1 serving of wet food.
Step-by-step explanation:
Plz mark Brainliest?
If a function is defined as
where both 
are continuous functions, then 
is also continuous where defined, i.e. where 
So, in your case, this function is continous everywhere, except where
To solve this equation, we can use the formula 
It means that, if the leading terms is 1, then the x coefficient is the opposite of the sum of the roots, and the constant term is the product of the roots.
So, we're looking for two terms whose sum is 7, and whose product is 12. These numbers are easily found to be 3 and 4.
So, this function is continuous for every real number different than 3 or 4.
