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vovangra [49]
3 years ago
15

Sketch the graphs of the line system on a corporate:{2x-y=1} {y=5x-5}

Mathematics
1 answer:
Papessa [141]3 years ago
8 0
I don’t know what some of that means but I can do the start for you? You rearrange
2x-y=1.

To do this, you +y to both sides, giving you 2x=1+y and then you minus 1, giving you 2x-1=y

Which can be rewritten the other way round to make it slightly easier

y=2x-1

You also have y=5x-5

These are both straight line equations and are now in the form y=mx+c

To sketch these graphs I would do two tables.

X -3 -2 -1 0 1 2 3
Y

For this, you now substitute each of the values for X into one of the equations you have. This is 2x-y=1 (2x-1=y)


X -3 -2 -1 0 1 2 3
Y -7 -5 -3 -1 1 3 5

You may have noticed a pattern there, the y values increased by two each time. This makes it linear. You would plot that line, onto an axis, using the coordinates you now have.
So, (-3, -7), (-2,-5), (-1,-3), (0,-1), (1,1), (2,3), (3,5)

Then I would do the same for the second equation, and plot that too.


X -3 -2 -1 0 1 2 3
Y -20 -15 -10 -5 0 5 10

You may have spotted this time the values increased by 5.

Then again plot this line using the coordinates shown.

I honestly have no idea what it means by “the line system on a corporate” but if that means on an axis then there’s your answer. If not then I do not know.

Hope this helps?
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Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m
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The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

8\sqrt{2}-6\sqrt{2}  =2\sqrt{2}

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

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