A graphing calculator shows it takes 3.25 seconds for the shoe to hit the building.
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200 -16t^2 = 31
169 = 16t^2
√(169/16) = t
t = 13/4 = 3 1/4
I think it would be 3569 , i’m not sure tho.
Answer:
∠J = 60°
Step-by-step explanation:
The Law of Cosines tells you ...
j² = k² +l² -2kl·cos(J)
Solving for J gives ...
J = arccos((k² +l² -j²)/(2kl))
J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)
J = 60°
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<em>Additional comment</em>
It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.
Answer:
Perimeter = 120m
Step-by-step explanation:
I'm going to assume that your function is f(x) = 1 + x^2 (NOT x2).
I suspect you're trying to estimate the "area under the curve of f(x) = 1 + x^2. You need to use this or a similar description to explain what you're doing.
Also, you need to specify whether you want "left end points" or "right end points" or "midpoints." Again I must assume you want one or the other (and will assume that you meant "left end points").
First, let's address the case n=3. You must graph f(x) = 1 + x^2 between -1 and +1. We will find the "lower sum," using "left end points." The 3 x-values are {-1, -1/3, 1/3}. Evaluate the function f(x) = 1 + x^2 at these 3 x-values. Keep in mind that the interval width is 2/3.
The function (y) values are {0, 2/3, 4/3}.
Sorry, Michael, but I must stop here and await clarification from you regarding what you've been told to do in this problem. Otherwise too much guessing (regarding what you meant) is necessary. Please review the original problem and ensure that you have copied it exactly as presented, and also please verify whether this problem does indeed involve estimating areas under curves between starting and ending x-values.