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?
1/2 of getting an even number and 1/6 of getting a 5.
Answer:
(0, 11) is a solution.
(8, 12) is a solution.
(16, 13) is a solution
(1, 89/8)
(2, 45/4)
Step-by-step explanation:
y = 1/8x + 11
for this equation....
y = (1/8) x + 11
let x = 0 get
y = 0 + 11 = 11
(0, 11) is a solution.
(8, 12) is a solution.
(16, 13) is a solution
(1, 89/8) because y = 1/8 + 11 = 1/8 + 88/8 = 89/8
(2, 45/4)
The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.
Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars 
(not all zero) such that
which leads to the system of linear equations,
From the third equation, we have 
, and substituting this into the second equation gives
and in turn, 
. Substituting these into the first equation gives
which tells us that any value of 
will work. If 
, then 
and 
. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.
Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.
