Graph the relation shown in the table. Is the relation a function? Why or why not?
2 answers:
Answer: choice B) yes; only one range value exists for each domain value
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Explanation:
The inputs are x = -3, x = -1, x = 1, x = 5. They are the first coordinate listed of each point. We don't have any x values repeating so this means we have a function. Each input leads to exactly one output which is what choice B is stating. The domain is the set of allowed inputs, or x values. The range is the set of possible y outputs.
If we had something like (1,2) and (1,5) then the input x = 1 leads to more than one output (y = 2 and y = 5). This example means we don't have a function
If you graph the points (-3, -2), (-1,0), (1,0) and (5,-2) as shown in the attached image, then you'll notice that it is impossible to pass a single line through more than one point. Therefore this graph passes the vertical line test visually proving we have a function.
Going back to the example with (1,2) and (1,5), plotting these two points leads to the vertical line test failing implying we don't have a function.
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---------------------------------
Explanation:
The inputs are x = -3, x = -1, x = 1, x = 5. They are the first coordinate listed of each point. We don't have any x values repeating so this means we have a function. Each input leads to exactly one output which is what choice B is stating. The domain is the set of allowed inputs, or x values. The range is the set of possible y outputs.
If we had something like (1,2) and (1,5) then the input x = 1 leads to more than one output (y = 2 and y = 5). This example means we don't have a function
If you graph the points (-3, -2), (-1,0), (1,0) and (5,-2) as shown in the attached image, then you'll notice that it is impossible to pass a single line through more than one point. Therefore this graph passes the vertical line test visually proving we have a function.
Going back to the example with (1,2) and (1,5), plotting these two points leads to the vertical line test failing implying we don't have a function.
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So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus![\bf \begin{cases} b=\textit{rate of the boat}\\ r=\textit{rate of the river} \end{cases}\qquad thus \\\\\\ \begin{array}{lccclll} &distance&rate&time(hrs)\\ &----&----&----\\ upstream&48&b-r&4\\ downstream&48&b+4&3 \end{array} \\\\\\ \begin{cases} 48=(b-r)(4)\to 48=4b-4r\\\\ \frac{48-4b}{-4}=r\\ --------------\\ 48=(b+r)(3)\\ -----------------------------\\\\ thus\\\\ 48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ab%3D%5Ctextit%7Brate%20of%20the%20boat%7D%5C%5C%0Ar%3D%5Ctextit%7Brate%20of%20the%20river%7D%0A%5Cend%7Bcases%7D%5Cqquad%20thus%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26distance%26rate%26time%28hrs%29%5C%5C%0A%26----%26----%26----%5C%5C%0Aupstream%2648%26b-r%264%5C%5C%0Adownstream%2648%26b%2B4%263%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0A48%3D%28b-r%29%284%29%5Cto%2048%3D4b-4r%5C%5C%5C%5C%0A%5Cfrac%7B48-4b%7D%7B-4%7D%3Dr%5C%5C%0A--------------%5C%5C%0A48%3D%28b%2Br%29%283%29%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Athus%5C%5C%5C%5C%0A48%3D%5Cleft%5B%20b%2B%5Cleft%28%5Cboxed%7B%5Cfrac%7B48-4b%7D%7B-4%7D%7D%5Cright%29%20%5Cright%5D%20%283%29%0A%5Cend%7Bcases%7D)
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"