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dusya [7]
3 years ago
15

What is -2,1 after a rotation 180 degrees clockwise around the origin

Mathematics
2 answers:
andrew-mc [135]3 years ago
8 0

Answer:

It would be 2,-1

Step-by-step explanation:

A 180 degree clockwise rotation is (x,y) going to (-x,-y)

MakcuM [25]3 years ago
6 0

Answer:

(2,1)

Step-by-step explanation:

The point (-2,1) is in the second quadrant.

After a 180° clockwise rotation, this point becomes (2,1) and in the first quadrant.

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Use the least common denominator to enter an equivalent fraction for each fraction.​
kykrilka [37]

Answer:

1/9 = 5/45 2/15 = 6/45

Step-by-step explanation:

9: 9, 18, 27, 36, 45

15: 15, 30, 45

5 0
2 years ago
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
OKAY THIS IS THREE QUESTIONS SO HELP PLS!
blondinia [14]
First one is c
second one is b
and third one is c
4 0
3 years ago
For each incorrect answer on a test, the teacher changes the total points possible by - 3
Bogdan [553]

Answer:

40

Step-by-step explanation:

8 incorrect answers x -3 points per wrong answer. 64 - 24 points taken off = score of 40.

5 0
3 years ago
Read 2 more answers
Solve for x.
Zigmanuir [339]
X/2 >= -4
x >= -4 * 2
x >= -8
6 0
3 years ago
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