Melody

g(x) = x^2 - 11x + 30

g(f(x)) = x^4 - 8x^3 + 11x^2 + 20x + 6

Let  f(x)= x^2 +Bx+C

then

g(f(x))= (x^2 +Bx+C)^2 - 11(x^2 +Bx+C) +30

Now just expeand it and equate the coefficients for find B and C ,

It is not that difficult.

Can you check your input please.

You have x^2 twice in the second function.

And welcome to the web2.0 forum.   I hope you have fun and learn lots.

Like this.

\(tan17=\frac{x}{y}\qquad (1)\\ tan 21.8 = \frac{x+8}{y}\qquad (2)\)

solve simultaneously.

Hint,

You cannot divide by 0

help you do what?

This is not difficult.

Draw a clock face and see when it is true.  The two hands must be at right angles to each other and since it is n o'clock the big hand must point to the 12.

3√6cis(π/8) and 2√5cis(7π/6)

assuming only the 5 sand 5 are under the square root

\([3\sqrt6 cis(π/8) ]*[ 2\sqrt5 cis(7π/6)]\\ =6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\ =6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\ =6\sqrt{30}*e^{(\pi/8)i}*e^{(7\pi/6)i}\\ =6\sqrt{30}*e^{((1/8)+(7/6))\pi i}\\ =6\sqrt{30}*e^{((3/24)+(28/24))\pi i}\\ =6\sqrt{30}*e^{(31/24)\pi i}\\ =6\sqrt{30}*cis{(31\pi/24)}\\ \)

I have not checked it.  That is your job

LaTex

[3\sqrt6 cis(π/8) ]*[ 2\sqrt5 cis(7π/6)]\\
=6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\
=6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\
=6\sqrt{30}*e^{(\pi/8)i}*e^{(7\pi/6)i}\\
=6\sqrt{30}*e^{((1/8)+(7/6))\pi i}\\
=6\sqrt{30}*e^{((3/24)+(28/24))\pi i}\\
=6\sqrt{30}*e^{(31/24)\pi i}\\
=6\sqrt{30}*cis{(31\pi/24)}\\

4!*8C4

Zep. looks like I did so i better add 6.  That makes   44  And now we agree.

Thanks guest  :))