Answer:
The length of a side of the square is 6.3 units.
Step-by-step explanation:
When given vertices for a given shape, the length of the side is calculated using the formula:
√(x2 - x1)² + (y2 - y1)²
When given vertices (x1 , y1) and (x2 , y2)
Square ABCD has vertices A(-2,-3), B(4, -1), C(2,5), and D(-4,3). Find the length of a side
Side AB : A(-2,-3), B(4, -1)
√(x2 - x1)² + (y2 - y1)²
= √(4 -(-2))² + (-1 -(-3))²
= √ 6² + 2²
= √36 + 4
= √40
= 6.3245553203
≈ 6.3 units
B(4, -1), C(2,5),
√(x2 - x1)² + (y2 - y1)²
= √ (2- 4)² + (5- (-1))²
= √-2² + 6²
= √4 + 36
= √40
= 6.3245553203
≈ 6.3 units
C(2,5), D(-4,3).
√(x2 - x1)² + (y2 - y1)²
= √(-4 - 2)² + (3 - 5)²
= √-6² + -2²
= √36 + 4
= √40
= 6.3245553203
≈ 6.3 units
A(-2,-3), D(-4,3).
√(x2 - x1)² + (y2 - y1)²
= √(-4 -(-2))² + (3 - (-3))²
= √-2² + 6²
= √4 + 36
= √40
= 6.3245553203
≈ 6.3 units
Answer: Your friends it farther away from school, because 1030 meters converted to km is 1.030 km which is smaller and closer than 1.5 km which further away
Step-by-step explanation: convert 1.030km to 1.5km
the correct answer is c 7 feet i hope this helps you out
Answer:
Step-by-step explanation:
If the position function is
and we are looking for time when the height is 0, we sub in a 0 for h(t) and solve for t:
and the easiest way to do this is to factor by taking the GCF of -40t:
0 = -40t(t - 5) and by the Zero Product Property,
-40t = 0 or t - 5 = 0. Solving for t, we get
t = 0 (which is before the object is launched) and
t = 5 (which is how long it takes the object to go from the ground, up to its max height, and then back to the ground again).
Your choice is c) 5