Answer:
t = 2.28 s
Step-by-step explanation:
h = 105 - 9t - 16t ^ 2
0 ft = 105 ft - 9t -16^t
To find the roots of a quadratic function we have to use the Bhaskara formula
, the roots will give us the time it takes to reach zero height
ax^2 + bx + c = 0
-16^t - 9t + 105 ft = 0 ft
a = -16 b = -9 c = 105
t1 = (-b + √ b^2 - 4ac)/2a
t2 =(-b - √ b^2 - 4ac)/2a
t1 = (9 + √(-9^2 - (4 * (-16) * 105)))/2 * (-16)
t1 = (9 + √(-81 + 6720))/ -32
t1 = (9 + √6639)/ -32
t1 = (9 + 81.84)/ -32
t1 = 90.84 / -32
t1 = -2.83 s
t2 = (9 - √(-9^2 - (4 * (-16) * 105)))/2 * (-16)
t2 = (9 - √(-81 + 6720))/ -32
t2 = (9 - √6639)/ -32
t2 = (9 - 81.84)/ -32
t2 = -72.84 / -32
t2 = 2.28 s
we have two possible values, we are only going to take the positive one, beacause we are talking about time
t2 = 2.28 s
Answer:
m = - 3
Step-by-step explanation:
a³ + 27 ← is a sum of cubes and factors in general as
a³ + b³ = (a + b)(a² - ab + b²), thus
a³ + 27
= a³ + 3³
= (a + 3)(a² - 3a + 9)
comparing a² - 3a + 9 to a² + ma + 9, then
m = - 3
This graph is not a minus graph so it just keeps going on and on to infinite.
I’ve attached a pic of the graph
You could do 5+10 or 12+3
Answer: s = 8
Step-by-step explanation:
