Answer:
Are you from us
Step-by-step explanation:
please I beg you
In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05
<h3>
Answer:</h3>
- a_n = -3a_(n-1); a_1 = 2
- a_n = 2·(-3)^(n-1)
<h3>
Step-by-step explanation:</h3>
A) The problem statement tells you it is a geometric sequence, so you know each term is some multiple of the one before. The first terms of the sequence are given, so you know the first term. The common ratio (the multiplier of interest) is the ratio of the second term to the first (or any term to the one before), -6/2 = -3.
So, the recursive definition is ...
... a_1 = 2
... a_n = -3·a_(n-1)
B) The explicit formula is, in general, ...
... a_n = a_1 · r^(n -1)
where r is the common ratio and a_1 is the first term. Filling in the known values, this is ...
... a_n = 2·(-3)^(n-1)
Answer:
3/4
Step-by-step explanation:
- cos (G) = adjacent/opposite
- 4 is the opposite side and 3 is the adjacent
