The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:
x + y = 3 and <span>2x + 2y = 6
Determinant of the equations are </span>
<span>| 1 1 | </span>
<span>| 2 2 | = 0
</span>
the numerator determinants would be
<span>| 3 1 | . .| 1 3 | </span>
<span>| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. F</span>or instance (3, 0), <span>(2, 1) and (4, -1). Therefore, it would have infinitely many solutions. </span>
Answer:
yes
Step-by-step explanation:
the second number in a set of parenthesis is the y value and the number in the parenthesis is 92 which makes y=92 true
<h3><u>Answer</u><u>:</u><u>-</u></h3>
x=17
<h3><u>step-by</u><u>-</u><u>step</u><u> </u><u>Explanation</u><u>:</u><u>-</u></h3>

<h3 />
This is a isosceles triangle. As it is a triangle we can apply sum theory. we have to take the sum of given unknown polynomials as 180° .Then by solving it we can find the value of x.
<h3><u>Solution</u><u>:</u><u>-</u></h3>
Given angles
According to sum theory



- Together like polynomials and constants







<span> -(x + 1) - 5 > 10
-x - 1 - 5 > 10
- x - 6 > 10
Add 6 to the both sides,
-x - 6 + 6 > 10 + 6
-x > 16
x < -16 [ changed sign 'cause changed -sign ]
In short, Your Answer would be x < -16
Hope this helps!</span>
I'd say yes. If you use the diagonal as a reference. Take the square and set your compass to the width of the diameter of the square. Now put it on the page and mark a point. Put the point of the compass on that mark and make another mark. Now you can connect the two marks with the straight edge and you have a line that, if you made a square with sides that long, it'd have 2x the area of the first one. That's because the diagonal is the square root of 2 larger than one side. Square the square root of 2 and you've got 2. You lust need to make a perpendicular line to the first one to get the box going.