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forsale [732]
3 years ago
14

Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D'E'F'G'H'I': Hexagon DEFGHI and Hexagon D prime E

prime F prime G prime H prime I prime on the coordinate plane with ordered pairs at D are 3, 5, at E 7, 5, at F 8, 2, at G 7, negative 1, at H 3, negative 1, at I 2, 2; at D prime negative 6, 2, at E prime negative 2, 2, at F prime negative 1, negative 1, at G prime negative 2, negative 4, at H prime negative 6, negative 4, at I prime negative 7, negative 1 Which rule represents the translation of hexagon DEFGHI to hexagon D'E'F'G'H'I'? (x, y)→(x − 9, y − 3) (x, y)→(x − 3, y − 9) (x, y)→(x + 3, y + 3) (x, y)→(x + 9, y + 9)
Mathematics
1 answer:
BigorU [14]3 years ago
8 0
We write the translation one by one:

D (3, 5)

D' (-6, 2)

That translates to: (x - 9, y - 3), because that is for passing from D to D': (3 - 9, 5 - 3) = (-6, 2).

Therefore if it is true for D and D' then, it has to be true for all others (you can check) for that rule to be true, so the correct answer is:

(x, y)→(x − 9, y − 3)
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Find the exact value of tan(165°) using a difference of two angles
Katyanochek1 [597]

Answer:  -2+\sqrt{3}

=========================================================

Work Shown:

Apply the following trig identity

\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\

Now let's rationalize the denominator

\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\

----------------------

As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as \sqrt{3}-2 but it's the same as above.

5 0
3 years ago
Find two consecutive whole numbers such that the sum of their squares is 221.
Law Incorporation [45]
n^2+(n+1)^2=2n^2+2n+1=221
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n^2+n-110=0
(n-10)(n+11)=0\implies n=10,n=-11

But since n must be a whole number, we ignore n=-11. So the two integers are n=10 and n+1=11.
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How would I solve this?
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First ln on both sides to bring down the powers.
Ln and e cancel each other out.
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