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: D) 10
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Explanation:
I'm assuming points M and N are midpoints of segments FD and FE respectively. If that's the case, then segment DE is twice as long compared to segment MN. We consider MN to be a midsegment.
So,
DE = 2*(MN)
3x-2 = 2*(x+4)
3x-2 = 2x+8
3x-2x = 8+2
x = 10
Answer:
3(pi)
Step-by-step explanation:
8.33 (repeating 33) is 8.333333...
3(pi) = 9.42477796...
17/2 = 8.5
sqrt(64) = 8
Answer: 3(pi)
Answer:
4,320
Step-by-step explanation:
7 is closer to 10 then 0. 4,310 + 10 = 4,320
B.
You would first have to add 9 to each side so that the equation equaled 0. Then you could pull these numbers.
