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scoray [572]
3 years ago
5

Suppose a figure is located in Quadrant I. Which of the following sequences will result in an image that is located in Quadrant

II?
A. rotate 360 degrees counterclockwise, then shift 1 unit up
B. rotate 180 degrees counterclockwise, then shift 1 unit down
C. rotate 90 degrees counterclockwise, then shift 1 unit up
D. rotate 270 degrees counterclockwise, then shift 1 unit down
Mathematics
1 answer:
vladimir2022 [97]3 years ago
4 0
The answer that results in an image located in Quadrant II is <span>C. rotate 90 degrees counterclockwise, then shift 1 unit up. This is because Quadrant II is located next (at the left) of Quadrant I. This means that a 90 degree rotation and a shift up are safe enough to estimate it to land on Quadrant II.</span>
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Suppose we choose x=1 and y=\dfrac12. Then

f(x-y)=\sqrt{f(xy)+1}\implies f\left(\dfrac12\right)=\sqrt{f\left(\dfrac12\right)+1}\implies f\left(\dfrac12\right)=\dfrac{1+\sqrt5}2


Now suppose we choose x,y such that

\begin{cases}x-y=\dfrac12\\\\xy=2009\end{cases}


where we pick the solution for this system such that x>y>0. Then we find

\dfrac{1+\sqrt5}2=\sqrt{f(2009)+1}\implies f(2009)=\dfrac{1+\sqrt5}2

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How long did it take a biker to travel 20 miles at 12mph
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Place the steps for finding f'(X) in the correct order.<br><br><br> Plz help <br> Thank you!
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Answer:

See below

Step-by-step explanation:

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\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}

Suppose we choose a path along the x-axis, so that y=0:

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