Half of the flowers are yellow tulips.
3 of the 4 columns are marked T for tulips.
Of those 3, 2 of them are yellow.
This shows us that 6 of the 12 boxes, 1/2, are yellow tulips.
add all the numbers without a minus.
now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.
let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.
![\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D3-i%5Cimplies%20%26x-3%2Bi%3D0%5C%5C%20x%3D3%2Bi%5Cimplies%20%26x-3-i%3D0%5C%5C%20x%3D4i%5Cimplies%20%26x-4i%3D0%5C%5C%20x%3D-4i%5Cimplies%20%26x%2B4i%3D0%20%5Cend%7Bcases%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28x-3%2Bi%29%28x-3-i%29%28x-4i%29%28x%2B4i%29%3D%5Cstackrel%7By%7D%7B0%7D%20%5C%5C%5B2em%5D%20%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B%28x-3%29%2Bi%5D%5B%28x-3%29-i%5D%7D%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5Bx-4i%5D%5Bx%2B4i%5D%7D%3D0)
![\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5B%28x-3%29%5E2-i%5E2%5D%5Bx%5E2-%284i%29%5E2%5D%3Dy%5Cimplies%20%5B%28x-3%29%5E2-%28-1%29%5D%5Bx%5E2-%284%5E2i%5E2%29%5D%3D0%20%5C%5C%5B2em%5D%20%5B%28x-3%29%5E2-%28-1%29%5D%5Bx%5E2-%2816%28-1%29%29%5D%3D0%5Cimplies%20%5B%28x-3%29%5E2%2B1%5D%5Bx%5E2%2B16%5D%3D0%20%5C%5C%5B2em%5D%20%5B%28x%5E2-6x%2B9%29%2B1%5D%5Bx%5E2%2B16%5D%3Dy%5Cimplies%20%28x%5E2-6x%2B10%29%28x%5E2%2B16%29%3D0%20%5C%5C%5C%5C%5C%5C%20x%5E4-6x%5E3%2B10x%5E2%2B16x%5E2-96x%2B160%3D0%20%5C%5C%5C%5C%5C%5C%20x%5E4-6x%5E3%2B26x%5E2-96x%2B160%3D0%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%204%7D%7D%7B4%28x%5E4-6x%5E3%2B26x%5E2-96x%2B160%29%3D4%280%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%204x%5E4-24x%5E3%2B104x%5E2-384x%2B640%3Dy~%5Chfill)
- We are to find the time (number of minutes) is would take for 23 mg of the substance to be remaining.
- The formula for time is written as:
t = [t1/2 x In(Nt/No)] / In 2
where:
t1/2 = Half life = 4 minutes
No = Initial quantity of the sample = 90 mg
Nt = Amount of the sample left = 23 mg
t = time elapsed = ?
Hence,
t = [4 x In (23/90)] / -In 2
t = 7.8731645610906 minutes
Approximately to the nearest hundredth = 7.87 minutes
Therefore, there will be 23mg of substance remaining after 7.87 minutes.
To learn more, visit the link below:
brainly.com/question/16624562
