Answer:
1st picture: (0,4)
The lines intersect at point (0,4).
2nd picture: Graph D
2x ≥ y - 1
2x - 5y ≤ 10
Set these inequalities up in standard form.
y ≤ 2x + 1
-5y ≤ 10 - 2x → y ≥ -2 + 2/5x → y ≥ 2/5x - 2
When you divide by a negative number, you switch the inequality sign.
Now you have:
y ≤ 2x + 1
y ≥ 2/5x - 2
Looking at the graphs, you first want to find the lines that intersect the y-axis at (0, 1) and (0, -2).
In this case, it is all of them.
Next, you would look at the shaded regions.
The first inequality says the values are less than or equal to. So you look for a shaded region below a line. The second inequality says the values are greater than or equal to. So you look for a shaded region above a line.
That would mean Graph B or D.
Then you look at the specific lines. You can see that the lower line is y ≥ 2/5x - 2. You need a shaded region above this line. You can see the above line is y ≤ 2x + 1. You need a shaded region below this line. That is Graph D.
Answer:
The Alans's average for the course is 84.5
Step-by-step explanation:
We are given
There are 4 tests, A Term paper and a Final examination.
Score = 92, 78, 82, 90.
Term paper score = 80
Final examination score = 86
Weighted mean = ∑ w.f/∑w
also the sum of all the weights is 100% = 1
weighted mean = 15%*92 + 15%*78+15%*82+15%*90+20%*80+20%*86/1
= 51.3 + 33.2
= 84.5
Therefore the Alans's average for the course is 84.5
12j + 12 + 3j or 15j + 12. This is because if you use Distributive Property to distribute the 3, then you multiply 3 by all the number inside the parentheses. I have two expressions because once you get get the answer after you use the property, then you can combine like terms, such as 12j and 3j which is 15j. Hope this helps!
-4x-y+6
you needed to just add like terms
(a) 0.50 per pound, you do 6 divided by 12
(b) 9.17 pounds per field, you do 55 divided by 6
