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rusak2 [61]
3 years ago
6

Alex is working to simplify

Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
7 0
What’s the problem.

Explanation:
stich3 [128]3 years ago
4 0

Answer:

is this all?

Step-by-step explanation:

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ANSWERRRR QUICKKKK PLEASEEEE
Brums [2.3K]

Answer: sqrt(2)/2  which is choice D

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

Explanation

(3pi/4) radians converts to 135 degrees after multiplying by the conversion factor (180/pi).

The angle 135 degrees is in quadrant 2. We subtract the angle 135 from 180 to find the reference angle

180-135 = 45

Then you can use a 45-45-90 triangle to determine that the ratio of opposite over hypotenuse is sqrt(2)/2

sine is positive in quadrant 2

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

Alternatively, you can use a unit circle. Refer to the diagram below. In red, I've circled the angle 3pi/4 radians. The terminal point for this angle has a y coordinate of sqrt(2)/2

Recall that y = sin(theta).

6 0
3 years ago
18. Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbreviated conj(z) or conj(a
AnnyKZ [126]

Question (18):

Answer: (A) All of the following

(I am using z^* in place of conj(z)

z*z^*=(a+ib)(a-ib)=a^2-(ib)^2= a^2+b^2=\mbox{modulus(z)}^2

z+z^*=a+ib+a-ib=2a\\z-z^*=a+ib-a+ib=2bi


Question (19):

Answer: (B) -7

modulus of -7 is sqrt(49)=7

the other choices are all < 7


5 0
3 years ago
Help me get the answer to this
Alchen [17]
The answer is 465 and you are welcome
7 0
3 years ago
Haaaaaaaaaaaaaaaaaaaaalllllllp plz
MA_775_DIABLO [31]

Answer:

square root of 166, 13, 14

Step-by-step explanation:

5 0
3 years ago
20 POINTS!!!
notka56 [123]

\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------

(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s

6 0
3 years ago
Read 2 more answers
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