The distance covered at time t is listed below:
Distance at 2 sec = d(2) = 64 feet
Distance at 4 sec = d(4) = 256 feet
Distance at 6 sec = d(6) = 576 feet
Distance at 8 sec = d(8) = 1024 feet
We are to find the average rate of change between 2 seconds and 6 seconds. The average rate of change will be:
![\frac{d(6)-d(2)}{6-2} \\ \\ = \frac{576-64}{4} \\ \\=128](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%286%29-d%282%29%7D%7B6-2%7D%20%5C%5C%20%5C%5C%20%3D%20%5Cfrac%7B576-64%7D%7B4%7D%20%5C%5C%20%5C%5C%3D128)
Therefore, the average rate of distance between 2 seconds and 6 seconds is 128 feet per second. This represents the speed of the object. So the speed of the object between 2 and 6 seconds was 128 feet per second.
The range of the function the as given in the equation is; -∞ ≤ y < ∞.
<h3>What is the range of the function?</h3>
It follows from the task content that the function given is; y=3√x+8.
On this note, it follows that the range of the function is the set of all possible outputs, y values and in this case is; -∞ ≤ y < ∞.
It therefore follows that the range of the function is as indicated above as all X-values (domain) must be positive and the square root of x yields a positive or negative real number.
Read more on range;
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Answer:
994.5
Step-by-step explanation:
First, you need to figure out how many people are actually attending the event. It says that there are 50% more so we are going to have to add 56% of 150 to 150.
150 x 0.56 = 84
150 + 84 = 234
Now, we have to multiply this by 4.25, because this is the cost of the food per person.
234 x 4.25 = 994.5