Answer:
<u>It</u><u> </u><u>is</u><u> </u><u>(</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>)</u><u>³</u><u> </u><u>-</u><u> </u><u>9</u><u>x</u><u>(</u><u>3</u><u> </u><u>-</u><u> </u><u>x</u><u>)</u>
Step-by-step explanation:
Express 27 in terms of cubes, 27 = 3³:

From trinomial expansion:

open first two brackets to get a quadratic equation:

expand further:

take y to be 3, then substitute:

Answer:
50
Step-by-step explanation:
n (n + 1) = 2550
n^2 + n = 2550
n = 50
n + 1 = 51
Im pretty sure that she has 12 cups.
Answer:
Following are the response to the given question:
Step-by-step explanation:
Move b a bit further if the angle between cd and ab changes when you move b if you want to make a perpendicular point to cd The angle BEC is 90 ° for making the AB line perpendicular to the line CD to transfer point B to the angle between the two lines.
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)