Answer:
The bottom left system of equations has no solutions.
Step-by-step explanation:
A solution to the system of equations is a point where the graphs of equations intersect, because the intersection point represents a point that is common to both equations; Therefore, a system that has no solutions, will have graphs that will never intersect.
Now, looking at the 4 graphs given, we see that the upper-left graph represents lines which intersect; therefore, this system has a solution.
The upper-right graph represents only one equation and not a system—it could well be that the red line hides another line that is exactly beneath it, in which case the system would have infinitely many solutions.
The bottom-left graph represents lines that are parallel—they are never going to meet; therefore, this system has no solutions.
The bottom-right graph represents lines that have a point of intersect, and therefore, have a solution.
Thus only the bottom-right graph has no solutions.
4 is not equivalent to 4% or 1/25
We want to use elimination to solve
y = 0 (1)
x + y = 40 (2)
Multiply (1) by -1 to eliminate y.
-y = 0 (3)
Add (2) and (3).
x + y + (-y) = 40 + 0
x = 40
Answer: Multiply by -1.
Answer:
3x +-4
Step-by-step explanation:
well since it is linear functions you would want the Mx+B equation to be present and since you have to find rise over run you would get 3/1 which is just 3 translated giving you a slope of 3 and your y-intercept is your B so in forth giving you the answer of 3x+ -4
plz mark brainliest
Answer:
nElements = 0;
for (int i = 0; i < a2d.length; i++)
nElements += a2d[i].length;
isRectangular = true;
for (int i = 1; i < a2d.length && isRectangular; i++)
if (a2d[i].length != a2d[0].length)
isRectangular = false;
Step-by-step explanation:
2 dimensional array is created and initialize element with 0. and the taking loop for 2 dimensional array .such as
n Elements = 0;
for (int i = 0; i < a2d.length; i++)
n Elements += a2d[i].length;
A boolean variable isRectangular and 2 dimensional array has been created if rectangle is true the take a2d.length and isRectangular less than i and intiliaze i with 1 then increment the i .It will give true value.
If rectangular is false then apply if statement and see if a2d[i] is not equal to a2d[0]
if (a2d[i].length != a2d[0].length)
isRectangular = false;
