firstly let's convert the mixed fraction to improper fraction, then hmmm let's see we have two denominators, 5 and 3, and their LCD will simply be 15, so we'll multiply both sides by that LCD to do away with the denominators, let's proceed,
![\bf \stackrel{mixed}{2\frac{1}{3}}\implies \cfrac{2\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{7}{3}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{z}{5}-4=\cfrac{7}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{15}}{15\left( \cfrac{z}{5}-4 \right)=15\left( \cfrac{7}{3} \right)}\implies 3z-60=35 \\\\\\ 3z=95\implies z=\cfrac{95}{3}\implies z = 31\frac{2}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B1%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%203%2B1%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B7%7D%7B3%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7Bz%7D%7B5%7D-4%3D%5Ccfrac%7B7%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B15%7D%7D%7B15%5Cleft%28%20%5Ccfrac%7Bz%7D%7B5%7D-4%20%5Cright%29%3D15%5Cleft%28%20%5Ccfrac%7B7%7D%7B3%7D%20%5Cright%29%7D%5Cimplies%203z-60%3D35%20%5C%5C%5C%5C%5C%5C%203z%3D95%5Cimplies%20z%3D%5Ccfrac%7B95%7D%7B3%7D%5Cimplies%20z%20%3D%2031%5Cfrac%7B2%7D%7B3%7D)
Answer:
sorry bro just wanted the points but thanks have a great day
Step-by-step explanation:
Calculate area of the white region inside circle.
Area of the two white triangles in top left and bottom right:

Area of the two white quarter circles in the bottom left and top right:

Total area of unshaded white region inside circle:

Area of entire circle including the white and shaded regions

Area of shaded region is area of entire circle - area of unshaded white region

First you draw threes cookies, then you give a cookie to each child, then with the last cookie you split it in half and they each have 1 1/2 cookie.
The equation is option number c because someth = smth