Answer:
d = 13 (distance between two points)
Step-by-step explanation:
(4, -2) and (-1, 10)
(X1, Y1) and (X2, y2)
distance between two points = d = sqrt (X2 - X1)² + (Y2 - Y1)²
d = √ (-1 - 4)² + (10 - (-2))²
d = √25 + 144
d = √169
d = 13 (distance between two points)
Answer:
(- 2, 4 )
Step-by-step explanation:
y = 2x + 8 → (1)
y = - 2x → (2)
substitute y = 2x + 8 into (2)
2x + 8 = - 2x ( add 2x to both sides )
4x + 8 = 0 ( subtract 8 from both sides )
4x = - 8 ( divide both sides by 4 )
x = - 2
substitute x = - 2 into (2)
y = - 2(- 2) = 4
solution is (- 2, 4 )
Answer:
y = - 16t² + 55.6t + 6
Step-by-step explanation:
Using y - y₀ = vt - 1/2gt² where g = 32 ft/s², and v the velocity of the football
So y = y₀ + vt - 1/2 × (32 ft/s²)t²
y = y₀ + vt - 16t² where y₀ = 6.5 ft
y = 6 + vt - 16t²
Now, when t = 3.5 s, that is the time the teammate catches the ball after the quarterback throws it, y = 5 ft. Substituting these into the equation, we have
5 = 6.5 + v(3.5 s) - 16(3.5 s)²
5 = 6.5 + 3.5v - 196
collecting like terms, we have
5 - 6.5 + 196 = 3.5v
194.5 = 3.5v
v = 194.5/3.5 = 55.57 ft/s ≅ 55.6 ft/s
So, substituting v into y, our quadratic model is
y = 6 + 55.6t - 16t²
re-arranging, we have
y = - 16t² + 55.6t + 6
The answer to your question is 134:)
