There are 2 tangent lines that pass through the point
![y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B%28-1%2B%5Csqrt%7B3%29%5E2%7D%20%7D%20%28x-1%29%2B2)
and
![y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B%28-1-%5Csqrt%7B3%29%5E2%7D%20%7D%20%28x-1%29%2B2)
Explanation:
Given:
![y=\frac{x}{x+1}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bx%7D%7Bx%2B1%7D)
The point-slope form of the equation of a line tells us that the form of the tangent lines must be:
![[1]](https://tex.z-dn.net/?f=%5B1%5D)
For the lines to be tangent to the curve, we must substitute the first derivative of the curve for
:
![\frac{dy}{dx} =\frac{d(x)}{dx}(x+1)-x^\frac{d(x+1)}{dx} \\ \\](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%5Cfrac%7Bd%28x%29%7D%7Bdx%7D%28x%2B1%29-x%5E%5Cfrac%7Bd%28x%2B1%29%7D%7Bdx%7D%20%5C%5C%20%5C%5C)
![\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%5Cfrac%7Bx%2B1-x%7D%7B%28x%2B1%29%5E2%7D)
![\frac{dy}{dx}= \frac{1}{(x+1)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D%20%5Cfrac%7B1%7D%7B%28x%2B1%29%5E2%7D)
![[2]](https://tex.z-dn.net/?f=%5B2%5D)
Substitute equation [2] into equation [1]:
![[1.1]](https://tex.z-dn.net/?f=%5B1.1%5D)
Because the line must touch the curve, we may substitute ![y=\frac{x}{x+1}:](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bx%7D%7Bx%2B1%7D%3A)
![\frac{x}{x+1}=\frac{x-1}{(x+1)^2}+2](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bx%2B1%7D%3D%5Cfrac%7Bx-1%7D%7B%28x%2B1%29%5E2%7D%2B2)
Solve for x:
![x(x+1)=(x-1)+2(x+1)^2](https://tex.z-dn.net/?f=x%28x%2B1%29%3D%28x-1%29%2B2%28x%2B1%29%5E2)
![x^2+x=x-1+2x^2+4x+2](https://tex.z-dn.net/?f=x%5E2%2Bx%3Dx-1%2B2x%5E2%2B4x%2B2)
![x^2+4x+1](https://tex.z-dn.net/?f=x%5E2%2B4x%2B1)
![x\frac{-4±\sqrt{4^2-4(1)(1)} }{2(1)}](https://tex.z-dn.net/?f=x%5Cfrac%7B-4%C2%B1%5Csqrt%7B4%5E2-4%281%29%281%29%7D%20%7D%7B2%281%29%7D)
± ![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
±
<em> </em>
![x=-2-\sqrt{3}](https://tex.z-dn.net/?f=x%3D-2-%5Csqrt%7B3%7D)
There are 2 tangent lines.
![y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B%28-1%2B%5Csqrt%7B3%29%5E2%7D%20%7D%20%28x-1%29%2B2)
and
![y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B%28-1-%5Csqrt%7B3%29%5E2%7D%20%7D%20%28x-1%29%2B2)
Answer:
Horizontal distance = 1218.41 ft
Step-by-step explanation:
Given data:
Slope distance = 1223.88 ft
Zenith angle is = 95°25'14"
converting zenith angle into degree
Zenith angle is
94.421°
Horizontal distance![= S\times sin(Z)](https://tex.z-dn.net/?f=%3D%20S%5Ctimes%20sin%28Z%29)
putting all value to get horizontal distance value
Horizontal distance ![= 1223.88\times sin(95.421)](https://tex.z-dn.net/?f=%3D%201223.88%5Ctimes%20sin%2895.421%29)
Horizontal distance = 1218.41 ft
Let's start by plugging in 1 for y. You get 7-2/6, which is 5/6. The other equation is 3-7/12, which equals -1/3. Now let's try 10. You get 7-2(10)/6, which is -13/6. The other equation is 3(10)-7/12, which is 23/12, and is larger!
1 is a value of y that could work!
Hope this helps!! :)
Answer:
there are no options listed, but some statements that are true about this question are:
- total perimeter = (8.5 x 2) + (5.5 x 2) = 28 feet = 9¹/₃ yards
- since Shelley purchased 10 yards of ribbon, it should be enough
- after finishing the blanket, Shelley will have 2 feet of ribbon left
2^2. 2 and 2. Because 2 and 2