Answer is- 824% all you have to do is move the decimal point two spaces to the right.
To solve this problem you must apply the proccedure shown below:
1- The problem says that you are swimming each 25 yard length 3 seconds faster than your personal best and you have to calculate the change in time (which you can call
) of your personal best after 200 yards.
2- To calculate this, you must multiply 200 yards by 3 seconds and divide by 25 yards, as following:

Therefore, the answer is: 24 seconds.
Answer:
3 and 9
if f(x)=x^2+13 and g(x)=12x-14
Step-by-step explanation:
So when we are looking for the intersection of two functions, we are trying to figure out when they are the same. When you think same, you should think equal (=).
So we want to find when f(x)=g(x) for x.
f(x)=g(x)

Let's get everything to one side.
Subtracting 12x and adding 14 to both sides.

I'm going to reorder the left hand side and also simplify the 13+14 part:

Now since the coefficent of x^2 is just 1 our job is to find two numbers that multiply to be 27 and add up to be -12.
Those numbers are -3 and -9 since -3(-9)=27 and -3+(-9)=-12.
So the factored form of our equation is

Since the product is 0, then at least one of the factors must be 0.
So we want to solve both x-3=0 and x-9=0.
x-3=0 can be solved by adding 3 on both sides. This gives us x=3.
x-9=9 can be solved by adding 9 on both sides. This gives us x=9.
The intersection of f and g happens at x=3 or x=9.
recall your d = rt, distance = rate * time.
let's say we have two trains, A and B, A is going at 85 mph and B at 65 mph.
they are 210 miles apart and moving toward each other, at some point they will meet, when that happens, the faster train A has covered say d miles, and the slower B has covered then the slack from 210 and d, namely 210 - d.
When both trains meet, A has covered more miles than B because A is faster, however the time both have been moving, is the same, say t hours.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{Train A}&d&85&t\\ \textit{Train B}&210-d&65&t \end{array}\\\\ \dotfill\\\\ \begin{cases} \boxed{d}=85t\\ 210-d=65t\\[-0.5em] \hrulefill\\ 210-\boxed{85t}=65t \end{cases} \\\\\\ 210=150t\implies \cfrac{210}{150}=t\implies \cfrac{7}{5}=t\implies \stackrel{\textit{one hour and 24 minutes}}{1\frac{2}{5}=t}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7BTrain%20A%7D%26d%2685%26t%5C%5C%20%5Ctextit%7BTrain%20B%7D%26210-d%2665%26t%20%5Cend%7Barray%7D%5C%5C%5C%5C%20%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D85t%5C%5C%20210-d%3D65t%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20210-%5Cboxed%7B85t%7D%3D65t%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20210%3D150t%5Cimplies%20%5Ccfrac%7B210%7D%7B150%7D%3Dt%5Cimplies%20%5Ccfrac%7B7%7D%7B5%7D%3Dt%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20hour%20and%2024%20minutes%7D%7D%7B1%5Cfrac%7B2%7D%7B5%7D%3Dt%7D%20)
You add 70 + 50 then you multiply it by 7.5 and you get, 900 gallons.