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)
So it gives you the diameter but you need the radius of the circle to find the area so to find the radius you just divide the diameter by two. the formula for the area of a circle is A=pi*r^2 so you just multiply 3.14 by 3 squared so you just put it in the calculator as A=3.14*3^2 and you get 28.26 hope this helped
1 hour = 60 minutes
13/60 = 0.2
answer: 0.2<span>m/min
</span>
The smallest y can be is y = -1. The largest it can be is y = 1. The value of y can be anything in between. The value of y is a real number.
Answer: Choice D) All real numbers between -1 and 1 including -1 and 1Note: we can write that as

which in interval notation would look like
![[-1, 1]](https://tex.z-dn.net/?f=%5B-1%2C%201%5D%20)
. The square brackets tell us to include the endpoints.