The function f(x) = (x)? If that is the function, then there is no retriction on the Domain?
Answer:
80 yd^2
Step-by-step explanation:
Since the formula for squares is s^2, we can use this to set up an equation, which is
s^2=400
If we square root this, we get
s=20
The perimeter formula for a square is 4s, so if we plug in, we end up with 80 yd^2
Answer:
B
Step-by-step explanation:
In a boxplot, when the median is in the middle of the box, or when there are equal proportions around the median, the distribution is symmetric. As is shown here, the median is closer to the left side of the box, making it positively skewed (the mean is greater than the median) and not symmetric.
The upper semicircle ABCD has radius R=2 and area
![A_u= \frac{1}{2} \pi R^2= \frac{1}{2} \pi\cdot 2^2=2\pi](https://tex.z-dn.net/?f=A_u%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%20R%5E2%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%5Ccdot%202%5E2%3D2%5Cpi)
and the lower semicircle AD has radius r=1 and area
![A_l= \frac{1}{2} \pi r^2= \frac{1}{2} \pi\cdot 1^2= \frac{1}{2}\pi](https://tex.z-dn.net/?f=A_l%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%20r%5E2%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%5Ccdot%201%5E2%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20)
. The whole fugure area is
![A=A_u+A_l=2\pi + \frac{1}{2}\pi= \frac{5}{2} \pi](https://tex.z-dn.net/?f=A%3DA_u%2BA_l%3D2%5Cpi%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%3D%20%5Cfrac%7B5%7D%7B2%7D%20%5Cpi)
.
The perimeter consists of upper semicircle length, lower semicircle length and CD length.
![L_u= \frac{1}{2} \cdot(2\pi R)=2\pi](https://tex.z-dn.net/?f=L_u%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%282%5Cpi%20R%29%3D2%5Cpi)
![L_l= \frac{1}{2} \cdot(2\pi r)=\pi](https://tex.z-dn.net/?f=L_l%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%282%5Cpi%20r%29%3D%5Cpi)
![L_{CD}=2](https://tex.z-dn.net/?f=L_%7BCD%7D%3D2)
and the perimeter
Answer:
The y-intercept is (0, 34).
I used Demos graphing calculator and included an image of what values I put in.