Answer:
D) 14 seconds
Step-by-step explanation:
First we will plug 500 in for y:
500 = -4.9t² + 120t
We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:
500-500 = -4.9t² + 120t - 500
0 = -4.9t²+120t-500
Our values for a, b and c are:
a = -4.9; b = 120; c = -500
We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.
The quadratic formula is:
Using our values for a, b and c,
The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is
19-5 = 14 seconds.
Answer:
81°
Step-by-step explanation:
Please refer to the attachment(s) for explanations
You use the arithmetic sequence formula and input the information given to you.
tn = a + (n-1)d
t(56) is what your looking for so don't worry about the tn.
a is your first term,
a = 15.
n is the position of the term you are looking for, n = 56.
And d is the common difference, you find this by taking t2 and subtracting t1. t2=18 and t1=15.
d = 18 - 15 = 3
Inputting it all into the formula you get,
t(56) = 15 + (56-1)(3)
term 56 = 180.
You use this formula to find any term in a sequence provided you are given enough info. You can also manipulate it if you are asked to find something else like the first term(a), common difference(d) or term position(n). It just depends on what the question is asking and what information you are given. :)
Hope this helps!
Answer:
No he is not lol 16÷4=4, To buy 10, 10×4=40. 40-35=5 he needs 5 dollars more
