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)
<em> </em><em>Sum </em><em>of </em><em>both </em><em>the </em><em>angle </em><em>will </em><em> </em><em>1</em><em>8</em><em>0</em><em> </em><em>because</em><em> of</em><em> </em><em>linear </em><em>pair</em>
now, <em>1</em><em>0</em><em>x</em><em> </em><em>-</em><em> </em><em>2</em><em>0</em><em> </em><em>+</em><em> </em><em>6</em><em>x</em><em> </em><em>+</em><em> </em><em>8</em><em> </em><em> </em><em>=</em><em>. </em><em>1</em><em>8</em><em>0</em><em> </em>
<em> </em><em> </em><em>1</em><em>6</em><em>x</em><em> </em><em>-</em><em> </em><em>1</em><em>2</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em>
<em>1</em><em>6</em><em>x</em><em> </em><em>=</em><em> </em><em>1</em><em>9</em><em>2</em><em> </em>
<em>x=</em><em> </em><em>1</em><em>9</em><em>2</em><em>/</em><em>1</em><em>6</em>
<em>x </em><em>=</em><em> </em><em>1</em><em>2</em>
<em>1</em><em>s</em><em>t</em><em> </em><em>angle </em><em>=</em><em> </em><em>1</em><em>0</em><em>0</em>
<em>2</em><em>n</em><em>d</em><em> </em><em>is </em><em>=</em><em> </em><em>8</em><em>0</em>
<em>Hope</em><em> it</em><em> helps</em><em> and</em><em> your</em><em> day</em><em> will</em><em> full</em><em> of</em><em> happiness</em>
Answer:
40
Step-by-step explanation:
Two ways we can solve this problem:
1. Graphically
2. Mathematically
Because you already solved it graphically, I will show you how to do so mathematically. Of course, graphically is much much easier and more efficient in this problem.
Let's break the problem down.
First, we are given a graph which contains a slope.
To find slope, we use the technique -> rise/run
Picking 2 obvious points from the graph, we can see
1st point -> (10, 20)
2nd point -> (40, 80)
Now, let's find the slope

Now, we have an equation y = 2x, where y = number of pies and x = cups of sugar
We want to find how many cups of sugar we need to bake 80 pies. Simply substitute 80 = number of pies = y
y = 2x -> 80 = 2x
Solving for x, divide both sides by 2
40 = x
We need 40 cups of sugar.
Answer:
Step-by-step explanation:
9. the set containing all objects or elements and of which all other sets are subsets.
10. complement --> the amount in only one of teh sets
intersection --> the amount in both sets
11. 14 or -14
13. 46 1/2
14. 2 31/120
15. 1100% profit
16. 1200 gm
17. cube root of 7?
18. 6000 per year
4 1/7 = (4*100) +(1/7 of 100)= (400+ 14.28)= 414.28
The answer is 414.28