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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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If f(x) = 9x10 tan−1x, find f '(x).
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Answer:

\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}

General Formulas and Concepts:

<u>Calculus</u>

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Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)  

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<u>Step 1: Define</u>

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  3. Basic Power Rule:                                                                                         \displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

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