How does the area of triangle RST compare to the area of triangle LMN? is 2 square units less than the The area of △ RST area of △ LMN The area of △ RST is equal to the area of △ LMN The area of △ RST is 2 square units greater than the area of △ LMN The area of △ RST is 4 square units greater than the area of △ LMN.Jun 25, 2021
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:
Step-by-step explanation:
Here are the steps to follow when solving absolute value inequalities:
Isolate the absolute value expression on the left side of the inequality.
If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions.
If your problem has a greater than sign (your problem now says that an absolute value is greater than a number), then set up an "or" compound inequality that looks like this:
(quantity inside absolute value) < -(number on other side)
OR
(quantity inside absolute value) > (number on other side)
The same setup is used for a ³ sign.
If your absolute value is less than a number, then set up a three-part compound inequality that looks like this:
-(number on other side) < (quantity inside absolute value) < (number on other side)
The same setup is used for a £ sign
Answer:
optionb
Step-by-step explanation:
