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
68 is rational
18/42 is rational
70.2.29 is rational
91.154389 is rational
5 is
Informally, the domain is the set of all possible elements in a set for which there is one and only one output. In interval notation, the domain is (-4, ∞). I am assuming it continues on infinitely to the right because of the arrow. The range is [1, <span>∞). Here, you choose the lowest value on the graph, up to the highest one.</span>
The first carpenter is 0.4464 of the way is ahead of the second carpenter
<u>Solution:</u>
Given, Two carpenters are building a fence
After 5 minutes, one carpenter is finished 4/7 of the way
Second carpenter finished 1/8 of the way
Now, let us find the work done by each carpenter
Then, extra work done by first carpenter over second carpenter = work done by first carpenter – work done by second carpenter
= 0.5714 – 0.125 = 0.4464
Hence, the 1st carpenter is 0.4464 of the way is ahead of the 2nd carpenter
It’s not exactly equal to one, but in many cases in math they ask for a rounded answer.