Answer:
see explanation
Step-by-step explanation:
the equation of a circle in standard form is
(x - a)² + (y - b)² = r²
where (a, b) are the coordinates of the centre and r is the radius
(x - 8)² + (y + 6)² = 13 is in this form
with centre = (8, - 6) and r = 
Answer:
5000
Step-by-step explanation:
The 5 is in the "thousand" place, so it would be 5000
First you want to figure out what exactly it is you are looking for. We are looking for "capital letters that have rotational symmetry but do not have line symmetry"
So:
1. Must have rotational symmetry.
This means that if we rotate the capital letter 180 deg, either clockwise or counterclockwise, it will still look the same
2. Must not have line symmetry.
If an object has line symmetry, it means that if you draw a line down the middle (in any way), it will be symmetrical on both sides. We need capital letters that do not fit that condition.
Now we look at all capital letters.
We find that H, I, N,O, S, X, and Z are all rotationally symmetrical. Think about it. If you rotate them, they still look the same.
But, we have to make sure they do not have line symmetry. If we draw a line right down the middle of H, I, O and X (**note, the have multiple lines of symmetry), they are symmetrical on both sides of the line.
Now we are left with N, S, and Z
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
Answer:
y=3x-2
Step-by-step explanation:
The formula for slope is y=mx=b
You can find the slope by finding two points that lie on the line (1,1) & (0,-2)
The formula for slope is 
You plug the values in and you get 
Simplify and you get 3
The slope is 3
The y-intercept (b) is -2
