Answer:
ST = 7.07 units
Step-by-step explanation:
* Lets explain how to find the length of a segment
- The length of a segment whose endpoints are 
and
can be founded by the rule of the distance

* Lets solve the problem
∵ The line segment is ST
∵ S is (-3 , 10)
∵ T is (-2 , 3)
- Assume that S is
and T is 
∴
and 
∴
and 
- By using the rule above
∴ 
∴ 
∴ 
∴ 
∴ ST = 7.07 units
We can tell from the data that there is a midpoint between the lowest and highest point of the clock, which is at a height of 9.5 feet.
Moreover, the lowest point occurs at 6 o clock, and the highest occurs at 12 o clock.
The amplitude of variation from the mid-point is 0.5 feet given by (10 - 9) / 2.
Finally, the time period for the equation is 12 hours. Thus, the answer is:
h = 0.5cos(πt/6) + 9.5, option B
so he's losing 4/10 of a Kg daily, how many times will that equal 3.2 Kgs? namely how many times does 4/10 go into 3.2?
well, let's firstly convert 3.2 to a fraction, and then divide.
![\bf \stackrel{\textit{1 decimal}}{3.\underline{2}}\implies \cfrac{32}{\underset{\textit{1 zero}}{1\underline{0}}}\implies \cfrac{16}{5} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B1%20decimal%7D%7D%7B3.%5Cunderline%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B32%7D%7B%5Cunderset%7B%5Ctextit%7B1%20zero%7D%7D%7B1%5Cunderline%7B0%7D%7D%7D%5Cimplies%20%5Ccfrac%7B16%7D%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \cfrac{16}{5}\div \cfrac{4}{10}\implies \cfrac{16}{5}\div \cfrac{2}{5}\implies \cfrac{\stackrel{8}{~~\begin{matrix} 16 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 8](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B16%7D%7B5%7D%5Cdiv%20%5Ccfrac%7B4%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B16%7D%7B5%7D%5Cdiv%20%5Ccfrac%7B2%7D%7B5%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7B8%7D%7B~~%5Cbegin%7Bmatrix%7D%2016%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7D%7B~~%5Cbegin%7Bmatrix%7D%205%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%205%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7B~~%5Cbegin%7Bmatrix%7D%202%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%5Cimplies%208)
Solution: The correct option is option D.
Explanation:
The end behavior is the behavior of function as the input of the function approaches to very large value and very small value.
If a function is f(x), then at
and
the value of f(x) approaches to either infinity or negative infinity which defines the end behavior.
From the given table it is noticed that the as x decreases and approaches to large negative values then the values of the function is also approaches to the large negative values. According to the table as
,
.
From the given table it is noticed that the as x increases and approaches to large positive values then the values of the function is approaches to the large negative values. According to the table as
,
.
Since only option D satisfy the both conditions, therefore the correct option is option D.