AB + CD = BC + DA
<span>Where A, B, C, and D are points on the quad, and AB is the length of the line A to B, etc. </span>
<span>Since you know three sides, you can easily solve for the fourth. </span>
Answer:
C
Step-by-step explanation:
sum of angles in tri are 180
180-35-36=109
Answer:
Step-by-step explanation:
5x - y = 3.......(1)
2y = 10x+2.....(2)
Rearranging (2)
-10x + 2y = 2......(3)
Multiply equation (1) by 2
10x - 2y = 6..........(4)
Adding (3) and (4)
-10x + 10x + 2y - 2y = 2 + 6
No solution since both x and y are eliminated.
The fraction is 36/42.
36 + 42 = 78
36/42 = 6/7
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.
