9514 1404 393
Answer:
y = -x +3
Step-by-step explanation:
The point-slope form can be a useful place to start.
y -k = m(x -h) . . . . . line with slope m through point (h, k)
You require the line ...
y -(-4) = -1(x -7)
y = -x +7 -4 . . . . . . . . eliminate parentheses, add -4
y = -x +3 . . . . . . . . . slope-intercept form
Step-by-step explanation:
Let, x^a =y......(1) and
y^b =z.....(2) and
z ^c =x......(3).
Now, using (1) in (2) we get,
x ^ab =z......(4).
Now, using (4) in (3) we get,
x ^abc =x
or, x ^abc =x ^1
or, abc=1.
Hope it will help :)
If solving for x:
Because both equations are equal to Y, you can set both equations equal to each other:
5x+1=4x+8
Then it's a matter of simplification (subtract the 4x to the left side and subtract the 1 over to the right):
x=7
If solving for Y:
Plug x=7 into either equation and solve:
Y=4(7)+8
Y=36
3x-x=10+2
3x-1x=12
2x=12
2÷2. & 12÷2 =6
x=6
let's first off notice that, on the 2), the sector is really half of the whole circle, and on 3) the sector is one quarter of the whole circle.
now, on 2) AB is the diameter of 4 units, therefore it has a radius of 2, or half that.
![\bf \boxed{2} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2 \end{cases}\implies A=\pi 2^2\implies A=4\pi \\\\\\ \stackrel{\textit{half of that}}{A=2\pi}\implies A=\stackrel{\textit{rounded up}}{A=6.3~ft^2} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%202%5E2%5Cimplies%20A%3D4%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%7D%7D%7BA%3D2%5Cpi%7D%5Cimplies%20A%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D6.3~ft%5E2%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \boxed{3} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=20 \end{cases}\implies A=\pi 20^2\implies A=400\pi \\\\\\ \stackrel{\textit{one quarter of that}}{A=100\pi }\implies \stackrel{\textit{rounded up}}{A=314.2~in^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D20%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2020%5E2%5Cimplies%20A%3D400%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7BA%3D100%5Cpi%20%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D314.2~in%5E2%7D)
