Answer:
Step-by-step explanation:
The graph shows the solution (-6,2)
i.e at x= -6 y=2
Analysis of each of the answers, since we can't write the equation of a straight line with only that information i.e the single point
Then,
Option 1
1. 2x - 3y = -6
x= -6 y=2
Then let insert x=-6 and y =2
2(-6)-3(2)
-12-6
-18.
Since -18 ≠ -6, then this is not the equation of the line and doesn't make up the system
Option 2
2. 4x - y = 26
Inserting x=-6 and y=2
4(-6)-(2)
-24-2
-26
Since -26 ≠ 26, then this is not the equation of the line and doesn't make up the system
Option 3
3. 3x + 2y = -14
Inserting x=-6 and y=2
3(-6)+2(2)
-18+4
-14
Since -14 ≠ -14 then this is the equation of the line and it make up the system.
Option 4
x-y = -2
Inserting x=-6 and y=2
(-6)-(2)
-6-2
-8
Since -8≠ -2, then this is not the equation of the line and doesn't make up the system
Option 5
5. x+y=-4
Inserting x=-6 and y=2
(-6)+(2)
-6+2
-4
Since -4 ≠ -4, then this is the equation of the line and it makes up the system.
Then, there are two option that make up the system
3. 3x + 2y = -14
And
5. x+y=-4
5/9 to chose odd card, 3/6 to roll even number
Kate-1145
Karina-1029
Jane-916
Answer:Definition area"? Do you mean the "natural domain" of the function- the region in which the formula is defined? In order that a number have a square root that number must be non-zero.
Step-by-step explanation:Here, we must have x−1x≥0.
If x is positive, multiplying both sides by x we have x2−1=(x−1)(x+1)≥0. In order for that to be true, both x- 1 and x+ 1 must have the same sign: either x-1> 0 and x+ 1> 0 or x- 1< 0 and x+ 1< 0. The first pair of inequalities is true for x> 1 and the second for x< -1. Since "x is positive", we must have x> 1.
If x is negative, multiplying both sides by x we have x2−1=(x−1)(x+1)≥0. In order for that to be true, x- 1 and x+ 1 must have opposite signs: x+ 1> 0 and x- 1< 0 or x- 1<0 and x- 1> 0. The first pair is true for −1≤0≤1. The second pair are never both true. Since "x is negative" we must have −1≤x≤0.
Of course, we also cannot divide by 0 so x= 0 is not in the domain. The domain is the union of the two separate sets:{x|−1≤x<0}∪{x|x>1}.
Hope That Helps!
I might be wrong, but I think it is 8.
I think it’s 8 because all side lengths are equal in a square, meaning if one side length is 8 all the other side lengths are going to be 8 also.
