anonymous
  • anonymous
Determine whether the origin is included in the shaded region and whether the shaded region is above or below the line for the graph of the following inequality: y > three fifthsx − 2 The origin is not included in the shaded region and the shaded area is above the line. The origin is not included in the shaded region and the shaded area is below the line. The origin is included in the shaded region and the shaded area is above the line. The origin is included in the shaded region and the shaded area is below the line.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@queen-of-tokyo
anonymous
  • anonymous
@WhateverYouSay
anonymous
  • anonymous
@sweetburger

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More answers

anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
@yomamabf
zepdrix
  • zepdrix
Hey there :) Let's answer the second part of the question first, because it's really simple.
zepdrix
  • zepdrix
When you have something like this: \(\large\rm y\lt \color{orangered}{2x+3}\) Think of it like this: \(\large\rm y\lt \color{orangered}{\text{the line}}\) The orange part represents the line. So in this example, we want all of the `y values` which are `less than` `the line`. So in this example we would shade `below` the line.
zepdrix
  • zepdrix
How bout in your problem? y is GREATER THAN some line. So where we gonna shade?
anonymous
  • anonymous
above the line correct? @zepdrix
zepdrix
  • zepdrix
k good :) that already narrows down our choices!
zepdrix
  • zepdrix
\[\large\rm y\gt\frac{3}{5}x-2\]Recall that the origin is the point (0,0). So let's just plug it in... and find out if the inequality holds true!
zepdrix
  • zepdrix
\[\large\rm 0\gt\frac{3}{5}\cdot0-2\]After you simplify this a bit, does the inequality hold true?
anonymous
  • anonymous
no
anonymous
  • anonymous
its the same right?
zepdrix
  • zepdrix
So after simplifying, we have, 0 > -2
zepdrix
  • zepdrix
Hmm that appears to hold true, ya? 0 is larger than -2
anonymous
  • anonymous
oh yes my bad i was looking at the wrong thing
zepdrix
  • zepdrix
Ok good! So we figured out that the point (0,0) DOES in fact satisfy our inequality. So it is included in the shaded area.
anonymous
  • anonymous
thanks could you help mw with one more
zepdrix
  • zepdrix
sure, i can try :)
anonymous
  • anonymous
Given the system of equations presented here: 3x + 5y = 29 x + 4y = 16 Which of the following actions creates an equivalent system such that, when combined with the other equation, one of the variables is eliminated? Multiply the second equation by −1 to get −x − 4y = −16 Multiply the second equation by −3 to get −3x − 12y = −48 Multiply the first equation by −1 to get −3x − 5y = −29 Multiply the first equation by −3 to get −9x − 15y = −87
zepdrix
  • zepdrix
You're trying to match up the coefficients on either the x's or the y's, but not both. If we wanted to match up the x's, maybe we would need some kind of 3. If we wanted to match up the y's, it would be a little more difficult. We would multiply the first equation by 4, and the second equation by -5. They would then both have coefficients of 20, one of them being negative. So they would combine nicely. They didn't do that here though. They went the easier route. They're matching up the x's somehow.
zepdrix
  • zepdrix
We want them to match. Notice the first equation already has a 3, so it doesn't need anything more.
zepdrix
  • zepdrix
So what do you wanna do to the second equation to match them up?
anonymous
  • anonymous
add a 1 to the x?
zepdrix
  • zepdrix
The first equation has 3x. The second equation has x. If we add 1 to x, hmm I'm not really sure what that will do for us :d We want the same number of x in both equations.
zepdrix
  • zepdrix
We don't want to add. We want to multiply that second equation by some number that will turn it from x to some kind of 3x.
anonymous
  • anonymous
you would multiply it by 3 then
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
Good good good, we would multiply the second equation 3! That would give us 3x. But in order for them to combine, and disappear, we want one of them to be negative. So let's instead multiply the second equation by negative 3.
zepdrix
  • zepdrix
boom bam! yay team!
anonymous
  • anonymous
so the answer is D yes? @zepdrix
zepdrix
  • zepdrix
Option D talks about changes to equation 1. We were making changes to equation 2.
anonymous
  • anonymous
oh ok i wasn't reading carefully but thank you for your help :) @zepdrix

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