![\large\underline{\sf{Solution-}}](https://tex.z-dn.net/?f=%5Clarge%5Cunderline%7B%5Csf%7BSolution-%7D%7D)
<u>Let us assume that:</u>
![\sf \longmapsto x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } }](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Csqrt%7B6%20%2B%20%20%5Csqrt%7B6%20%2B%20%20%5Csqrt%7B6%20%2B%20...%20%5Cinfty%20%7D%20%7D%20%7D%20)
We can also write it as:
![\sf \longmapsto x = \sqrt{6 + x }](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Csqrt%7B6%20%2B%20%20x%20%7D%20)
Squaring both sides, we get:
![\sf \longmapsto {x}^{2} =6 + x](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%20%7Bx%7D%5E%7B2%7D%20%20%3D6%20%2B%20%20x)
![\sf \longmapsto {x}^{2} - x - 6 =0](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%20%7Bx%7D%5E%7B2%7D%20-%20x%20-%206%20%20%3D0)
By splitting the middle term:
![\sf \longmapsto {x}^{2} - 3x + 2x - 6 =0](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%20%7Bx%7D%5E%7B2%7D%20-%203x%20%2B%202x%20-%206%20%20%3D0)
![\sf \longmapsto x(x - 3) + 2(x - 3 ) =0](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%28x%20-%203%29%20%2B%202%28x%20-%203%20%29%20%3D0)
![\sf \longmapsto (x+ 2)(x - 3 ) =0](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%28x%2B%202%29%28x%20-%203%20%29%20%3D0)
<u>Therefore:</u>
![\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}](https://tex.z-dn.net/?f=%20%5Clongmapsto%5Cbegin%7Bcases%7D%20%5Csf%20%28x%2B%202%29%20%3D0%20%5C%5C%20%5Csf%20%28x%20-%203%29%20%3D%200%20%5Cend%7Bcases%7D)
![\sf \longmapsto x = - 2,3](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20-%202%2C3)
<u>But x cannot be negative. </u>
![\sf \longmapsto x = 3](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%203)
Therefore, the value of the expression is 3.
![\large\underline{\sf{Verification-}}](https://tex.z-dn.net/?f=%5Clarge%5Cunderline%7B%5Csf%7BVerification-%7D%7D)
Given:
![\sf\longmapsto x=3](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%3D3)
We can also write it as:
![\sf\longmapsto x = \sqrt{9}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B9%7D)
![\sf\longmapsto x = \sqrt{6+3}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B6%2B3%7D)
![\sf\longmapsto x = \sqrt{6 + \sqrt{9}}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B6%20%2B%20%5Csqrt%7B9%7D%7D)
![\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B6%20%2B%20%5Csqrt%7B6%2B3%7D%7D)
![\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B6%20%2B%20%5Csqrt%7B6%2B%5Csqrt%7B9%7D%7D%7D)
![\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B6%20%2B%20%5Csqrt%7B6%2B%5Csqrt%7B6%2B3%7D%7D%7D)
This pattern will continue.
![\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}](https://tex.z-dn.net/?f=%5Csf%5Clongmapsto%20x%20%3D%20%5Csqrt%7B6%20%2B%20%5Csqrt%7B6%2B%5Csqrt%7B6%2B...%5Cinfty%7D%7D%7D)
Well since she's trying to get 2 trays of equal cookies you subtract like how much does 27 go into 36 once so that's how much on each tray 27
Ok.... so the first part is asking you what the perimeter of the picture itself is.
The picture is 5in by 7in.... SO you add 5+5+7+7=24.... So the perimeter of the pic is 24 inches.
Now part b
So the frame is one inch thick
This mean that the sides are now 2 inches longeron each side
SO we have 7+7+9+9=32
So the perimeter is 32
I hope that helped!!!!! If you dont understand something ask and i will reexplain ;)
Answer:
for me the answer is letter B.