Set up an equation and solve:
2x-14=52
2x=66
x=33
sydney is 33 and devaughn is 19
Answer:
The Depth of the lake had increased by 19%.
Step-by-step explanation:
Given:
Depth of lake few months ago = 1300 ft
depth of lake currently = 1547 ft
We need to find the percent of increase in depth of lake.
Solution:
First we will find the increase in depth of lake.
Increase in depth of lake can be calculated by subtracting Depth of lake few months ago from depth of lake currently.
framing in equation form we get;
increase in depth of lake = 
Now to find the percent of increase in depth of lake we will divide increase in depth of lake from Depth of lake few months ago and then multiply by 100.
framing in equation form we get;
percent of increase in depth of lake = 
Hence the Depth of the lake had increased by 19%.
Answer:
1st, 2nd, and last.
Step-by-step explanation:
For the first two common rules can be applied such as 
= ![\frac{\sqrt[n]{a} }{\sqrt[n]{b} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5Bn%5D%7Ba%7D%20%7D%7B%5Csqrt%5Bn%5D%7Bb%7D%20%7D)
and that ![\sqrt[n]{\frac{a}{b} } = (\frac{a}{b})^{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7B%5Cfrac%7Ba%7D%7Bb%7D%20%7D%20%3D%20%28%5Cfrac%7Ba%7D%7Bb%7D%29%5E%7Bn%7D)
and for the last one this is just simplifying the radical and if you factor it you realize that, 
meaning that you can take the radical off of 5/8 and end up with that final answer
Dale drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 7 hours. when dale drove home, there was no traffic and the trip only took 5 hours. if his average rate was 18 miles per hour faster on the trip home, how far away does dale live from the mountains? do not do any rounding.
Answer:
Dale live 315 miles from the mountains
Step-by-step explanation:
Let y be the speed of Dale to the mountains
Time taken by Dale to the mountains=7 hrs
Therefore distance covered by dale to the mountain = speed × time = 7y ......eqn 1
Time taken by Dale back home = 5hours
Since it speed increased by 18 miles per hour back home it speed = y+18
So distance traveled home =speed × time = (y+18)5 ...... eqn 2
Since distance cover is same in both the eqn 1 and eqn 2.
Eqn 1 = eqn 2
7y = (y+18)5
7y = 5y + 90
7y - 5y = 90 (collection like terms)
2y = 90
Y = 45
Substitute for y in eqn 1 to get distance away from mountain
= 7y eqn 1
= 7×45
= 315 miles.
∴ Dale leave 315 miles from the mountains
