Answer:
t = 5.25 seconds
Step-by-step explanation:
Given that,
A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation:
It is assumed to find the time when the projectile will hit the ground. When the projectile hit the ground, its height is equal to 0 such that,
It forms a quadratic equation such that,
So, the projectile will hit the ground at t = 5.25 seconds.
A) multiplier = 1.13
30 x 1.13 = 33.9
B) multiplier = 0.85
250 x 0.8 = 200
C) multiplier = 0.72
825 x 0.72 = 594
Answer:
-1/16
Step-by-step explanation:
Answer:
a = - 2, b = - 6
Step-by-step explanation:
Substitute the values of the zeros into the polynomial and equate to zero.
x² +(a + 1)x + b
x = - 2 → (- 2)² - 2(a + 1) + b = 0 , that is
4 - 2a - 2 + b = 0
2 - 2a + b = 0 ( subtract 2 from both sides )
- 2a + b = - 2 → (1)
x = 3 → 3² + 3(a + 1) + b = 0, that is
9 + 3a + 3 + b = 0
12 + 3a + b = 0 ( subtract 12 from both sides )
3a + b = - 12 → (2)
Subtract (1) from (2) term by term to eliminate b
5a = - 10 ( divide both sides by 5 )
a = - 2
Substitute a = - 2 into either of the 2 equations and evaluate for b
Substituting into (2)
3(- 2) + b = - 12
- 6 + b = - 12 ( add 6 to both sides )
b = - 6
Thus a = - 2 and b = - 6
