Answer:

Step-by-step explanation:
We have to determine which trigonometric ratio to use, depending on what is displayed for us, and we will be using the <em>cotangent</em><em> </em>[or <em>tangent</em>] ratio, however, aside we are solving for an <em>angle</em><em> </em><em>measure</em>, we must use the inverse:

Extended Information on Trigonometric Ratios

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H(t) = -16t² + 16 t + 480
Factor out -16
-16(t² - t - 30)
t² - t - 30
(t + 5)(t - 6) ⇒ t(t-6) + 5(t-6) ⇒ t² -6t + 5t - 30 ⇒ t² - t - 30
t + 5 = 0 t - 6 = 0
t = -5 t = 6
The value of t is either -5 or 6 however time can't be measured in negative form; then, it took Rose 6 seconds to hit the water.
We have a "rectangular" double loop, meaning that both loops go to completion.
So there are 3*4=12 executions of t:=t+ij.
Assuming two operatiions per execution of the innermost loop, (i.e. ignoring the implied additions in increment of subscripts), we have 12*2=24 operations in all.
Here the number of operations (+ or *) is exactly known (=24).
Big-O estimates are used for cases with a varying scale of operations, governed by a variable (usually n) to indicate the sensitivity of the number of operations relative to a change in the size of n.
Here we do not have a scale, nor n is defined. The number of operations is constant and known at 24. So a variable is required to find the big-O estimate.
It’s c, for any integer,n
Answer:
1.101, 1.001, and 0.113
Step-by-step explanation:
the first on is the highest
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