Answer:
2.5, -2.61, 0.81, -0.12
Step-by-step explanation:
The taylor series of the function sin(x) around zero is given by
Therefore,
![\sin(\frac{5}{2})=\dfrac{5}{2}-\dfrac{[\frac{5}{2}]^3}{3!}+\dfrac{[\frac{5}{2}]^5}{5!}-\dfrac{[\frac{5}{2}]^7}{7!}+...](https://tex.z-dn.net/?f=%5Csin%28%5Cfrac%7B5%7D%7B2%7D%29%3D%5Cdfrac%7B5%7D%7B2%7D-%5Cdfrac%7B%5B%5Cfrac%7B5%7D%7B2%7D%5D%5E3%7D%7B3%21%7D%2B%5Cdfrac%7B%5B%5Cfrac%7B5%7D%7B2%7D%5D%5E5%7D%7B5%21%7D-%5Cdfrac%7B%5B%5Cfrac%7B5%7D%7B2%7D%5D%5E7%7D%7B7%21%7D%2B...)
hence the first four nonzero terms of the series are
![\dfrac{5}{2}=2.5\\\\-\dfrac{[\frac{5}{2}]^3}{3!} \approx -2.61\\\\\dfrac{[\frac{5}{2}]^5}{5!} \approx 0.81\\\\-\dfrac{[\frac{5}{2}]^7}{7!} \approx -0.12](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B2%7D%3D2.5%5C%5C%5C%5C-%5Cdfrac%7B%5B%5Cfrac%7B5%7D%7B2%7D%5D%5E3%7D%7B3%21%7D%20%5Capprox%20-2.61%5C%5C%5C%5C%5Cdfrac%7B%5B%5Cfrac%7B5%7D%7B2%7D%5D%5E5%7D%7B5%21%7D%20%5Capprox%200.81%5C%5C%5C%5C-%5Cdfrac%7B%5B%5Cfrac%7B5%7D%7B2%7D%5D%5E7%7D%7B7%21%7D%20%5Capprox%20-0.12)
Answer:
1 and 37/8 feet
Step-by-step explanation:
We set h(t) to 76 because we want to know when the ball will be at 76 feet high.
So, we have 76=-16t^2+90t+2. When we subtract 76 from both sides, we get -16t^2+90t-74=0, and when you mutiply the equation by -1 (so that x^2 has a positive coefficient, which makes things easier), we get 16t^2-90t+74=0.
We notice that each separate monomial has a common factor, 2, so we can take that out and make it 2(8t^2-45t+37) = 0. Factoring the quantity in the parentheses gives us 2(t-1)(8t-37) = 0.
Using Zero Product Property, we can set the two factors in parentheses to 0 (the 2 isn't really relevant so we don't need to worry about it). This gives us t-1=0 and 8t-37=0. Solving these two equations gives us t = 1 foot or t = 37/8 feet.
Answer:
Following are the response to the given points:
Step-by-step explanation:
For question 5.11:
For point a:
For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.
For point b:
The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last 
hours. Another sampling approach is a group.
For question 5.12:
This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a 
line graph when 5 parts were achieved.
This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours . An alternative type of analysis should be a random sample of five consecutive pieces created every hour.
To solve this problem, we need to use our knowledge that parallel lines have the same slope. If we have a line that is parallel to the line y = 2x - 7, our line must also have a slope of 2 (we know this line has a slope of 2 because it is in slope-intercept form, y=mx + b, where m represents slope and b represents the y-intercept).
Now, we can use our known slope and our given point to substitute into point-slope form, as shown below:
y = m (x - h) + k
y = 2 (x - 5) - 1
Next, we can simplify our equation by using the distributive property to get rid of the parentheses.
y = 2x - 10 - 1
Finally, we can complete our equation by computing the subtraction.
y = 2x - 11
We can conclude that this is the correct answer because if you plug in the point (5, -1) you get a true statement and it has a slope of 2.
Therefore, the answer is y = 2x - 11.
Hope this helps!
