Two points are 10 units apart on the coordinate plane. if
an ant were to walk between the 2 points moving only
vertically and horizontally, it would have to walk 6 horizontal
units. how many vertical units would the ant have to walk?

The vertical distance the ant have to walk is 8 units

Step-by-step explanation:

∵ The ant must be walk vertically and horizontally

∴ The distance from start point to end point will be as

   a hypotenuse of a right angle triangle

∵ The ant would have to walk 6 units horizontally that

   means one side of the right angle is 6 units

∵ The two points are apart by 10 units that means

   the length of the hypotenuse is 10 units

∴ The vertical distance the ant would have to walk is

   the length of the second side of the right angle

* By using Pythagoras Theorem:

∵ (hypotenuse)² = (horizontal leg)² + (vertical leg)²

∴ 10² = 6² + (vertical leg)²

∵ (vertical leg)² = 100 - 36 = 64

∴ Vertical leg  = √64 = 8 units

∴ The vertical distance the ant have to walk is 8 units

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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.