The position of an object in a video game is represented by an ordered pair. The coordinates of the ordered pair give the number of pixels horizontally and vertically from the origin. Use this scenario for Exercise.

a. Suppose that player A is located at (36, 315) and player B is located at (410, 53). How far apart are the players? Round to

the nearest pixel

b. If the two players move directly toward each other at the same speed, where will they meet?

c. If player A moves three times faster than player B, where will they meet? Round to the nearest pixel.

a. 457 pixels b. 229 pixels c. 114 pixels

Step-by-step explanation:

a. Using d = √[(x₂ - x₁)² + (y₂ - y₁)²] to find the distance apart of the coordinates (x₁, y₁) and (x₂, y₂) where (x₁, y₁) = (36, 315) and (x₂, y₂) = (410, 53).

So, d = √[(x₂ - x₁)² + (y₂ - y₁)²]

= √[(410 - 36)² + (53 - 315)²]

= √[374² + 262²]

= √[139,876 + 68,644]

= √208,520

= 456.64

= 457 pixels to the nearest pixel.

b. Since the players are d' distance apart, and moving at a same speed of v, if they approach each other, they meet at time t. So player A covers a distance of d = vt. For them to meet, player B approaches and covers a distance of d = d' - vt. So they meet when vt = d' - vt.

Solving this, we have

vt = d' - vt

vt + vt = d'

2vt = d'

t = d'/2v

Substituting t into d = vt, we have

d = vt

= v(d'/2v)

= d'/2

= 457/2

= 228.5

≅ 229 pixels  to the nearest pixel

c. Since the players are d' distance apart, and player A moves at a speed three times that of player B, if player B moves with speed v, then player A moves with speed 3v, as they approach each other, they meet at time t. So player A covers a distance of d = 3vt. For them to meet, player B approaches and covers a distance of d = d' - vt. So they meet when 3vt = d' - vt.

Solving this, we have

3vt = d' - vt

3vt + vt = d'

4vt = d'

t = d'/4v

Substituting t into d = vt, we have

d = vt

= v(d'/4v)

= d'/4

= 457/4

= 114.25

≅ 114 pixels to the nearest pixel

false

step-by-step explanation:

false.   two legs of the trapezoid would be congruent, but the other two would not.   were all the legs congruent, then we'd have a square; were the 2 shorter sides congruent and the 2 longer sides also congruent, we'd have a rectangle.