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AveGali [126]
3 years ago
5

50 POINTS PLEASE HELP Segment JK has endpoint J(2,4) and K(6,2). You dilate the segment using a dilation centered at the origin

with a scale factor of 1/2 and then reflect the image in the x-axis. What is the midpoint of the final image?
A (2, -1.5)
B (-2, 1.5)
C (2, 1.5)
D (1.5, 2)
Mathematics
1 answer:
stira [4]3 years ago
7 0

Answer:

C.

Step-by-step explanation:

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Find the nth term for the sequence 1, 4, 9, 16, 25, 36
gayaneshka [121]
1, 4, 9, 16, 25, 36
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If →u and →v are the vectors below, find the vector →w whose tail is at the point halfway from the tip of →v to the tip of →v−→u
S_A_V [24]
Û = (-1, -1, -1)
^v = (2, 3, -5)
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The required vector ^w = ((3/2 - 1/2), (2 - 1/2), (-2 - 1/2)) = (1, 1/2, -5/2)
5 0
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Find the values of cosΘ and tanΘ, given that sinΘ = 8/9 and Θ is in quadrant I
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sin^2 \alpha +cos^2 \alpha =1\ \ \ and\ \ \ tan \alpha = \frac{\big{sin \alpha }}{\big{cos \alpha }} \\\\sin \alpha  =  \frac{8}{9}\\\\ \Rightarrow\ \ \ (\frac{8}{9})^2+cos^2 \alpha =1\ \ \ \Rightarrow\ \ \ cos^2 \alpha =1- \frac{64}{81} \ \ \ \Rightarrow\ \ \ cos^2 \alpha = \frac{17}{81} \\\\ \alpha \ \in\ (0^0;90^0)\ \ \ \Rightarrow\ \ \ cos \alpha >0\ \ \ \Rightarrow\ \ \ cos \alpha = \frac{ \sqrt{17} }{9} \\\\

tan \alpha = \frac{8}{9}: \frac{ \sqrt{17} }{9} =\frac{8}{9}\cdot  \frac{9}{\sqrt{17}}= \frac{8}{\sqrt{17}}= \frac{8\cdot \sqrt{17}}{\sqrt{17}\cdot \sqrt{17}}=\frac{8\cdot \sqrt{17}}{17}
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