Problem 4:
![\rm \frac8{11}(n-10)=64](https://tex.z-dn.net/?f=%5Crm%20%5Cfrac8%7B11%7D%28n-10%29%3D64)
multiply by 11,
8(n-10) = 64*11
divide by 8,
n-10 = 8*11
Add 10,
n = 88 + 10 = 98
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Problem 5:
-0.6(r+0.2)=1.8
Divide by -0.6,
r+0.2 = -3
subtract 0.2,
r = -3 - 0.2 = -3.2
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Problem 6:
![\rm \left(w-\frac49\right)\left(-\frac23\right)=-\frac45](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%28w-%5Cfrac49%5Cright%29%5Cleft%28-%5Cfrac23%5Cright%29%3D-%5Cfrac45)
I hope this method isn't too confusing for you,
but I would multiply through by the LCM of the denominators to get rid of the fractions. LCM of 3, 9 and 5 is 45.
Multiply both sides by 45,
(45w - 20)(-30) = -36
Divide -30,
45w - 20 = 36/30
Add 20,
45w = 36/30 + 20
Divide 45 as a final step,
w = 21.2/45
You always get from one term to the next by adding 6:
![573 + 6 = 579](https://tex.z-dn.net/?f=573%20%2B%206%20%3D%20579)
![579 + 6 = 585](https://tex.z-dn.net/?f=579%20%2B%206%20%3D%20585)
![585 + 6 = 591](https://tex.z-dn.net/?f=585%20%2B%206%20%3D%20591)
![591 + 6 = 597](https://tex.z-dn.net/?f=591%20%2B%206%20%3D%20597)
So, the rule is "add 6"
The answer is the last option.
if she doesnt have the money ot begin with then the number is going to be negative which means less than ZERO.
V= πr2h
V= (3.14x12mx12m)8m
V= (452.16mx8m)
V=3,617.28m^3