Is each biconditional statement true or false? select true or false for each statement. true false a polygon is a square if and only if the polygon has exactly four sides. a polygon has exactly five sides if and only if the polygon is a pentagon. a polygon has exactly four sides if and only if the polygon is a rectangle.

Statement 1: A polygon is a square if and only if the polygon has exactly four sides. FALSE

Statement 2: A polygon has exactly five sides if and only if the polygon is a pentagon. TRUE

Statement 3:  A polygon has exactly four sides if and only if the polygon is a rectangle. FALSE

Step-by-step explanation:

In order to prove if a biconditional statement is true if both the conditionals are true, otherwise is false.

In this case you have to use geometry shapes definitions to determine if the conditionals are true or false.

The statement 1 is false because by definition a polygon is a square if the polygon has four sides of the same size forming rect angles. A polygon with four sides is not necessary a square.

The statement 2 is true because by definition a pentagon is a polygon with five sides .

The statement 3 is false because there are differents polygons with four sides (quadrilaterals). For example a square, rhombus. Not all the quatrilaterals are rectangles.

Statement 1: A polygon is a square if and only if the polygon has exactly four sides. FALSE

Statement 2: A polygon has exactly five sides if and only if the polygon is a pentagon. TRUE

Statement 3:  A polygon has exactly four sides if and only if the polygon is a rectangle. FALSE

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Step-by-step explanation:

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.