hectictar

Let's say the unknown coefficient of the linear term is  b , so the quadratic expression is:

4x²  +  bx  -  24

And we want to know what  b  is.

Since we know that one of the roots is  4,  we know that if we plug in  4  for  x  into the expression, it will equal  0

So we can make the following equation:

4(4)²  +  b(4)  -  24  =  0

Can you solve the above equation for  b  ?

I'm not totally sure, but here's my guess:

Sequence A is:   30, 48, 66, 84, 102, 120, . . .

Sequence B is:   30, 12, -6, -24, -42, -60, . . .

So we can say

A₁   =   30   =   30 + 18*0

A₂   =   48   =   30 + 18*1

A₃   =   66   =   30 + 18*2

A₄   =   84   =   30 + 18*3

A_n   =   30 + 18*(n - 1)

B₁   =   30   =   30 - 18*0

B₂   =   12   =   30 - 18*1

B₃   =   -6   =   30 - 18*2

B_n   =   30 - 18*(n - 1)

And so..

The 51st term of sequence A   =   A₅₁   =   30 + 18*(51 - 1)   =   30 + 18*50   =   930

The 51st term of sequence B   =   B₅₁   =   30 - 18*(51 - 1)   =   30 - 18*50   =   -870

The absolute value of the difference between  A₅₁  and  B₅₁   =   | 930 - -870 |   =   1800

Here is a balloon for the party!!

0.1x  -  1.6   =   0.2x  +  2.3

0.1x  -  0.2x   =   2.3  +  1.6

-0.1x  =  3.9

x  =  -39

By the angle sum/difference formula for cos,

cos(a - b)  =  cos a cos b + sin a sin b

And so...

\({\cos(\frac{\pi}{6}-x)}\ =\ \cos(\frac{\pi}{6})\cos(x)+\sin(\frac{\pi}{6})\sin(x)\\~\\ \phantom{\cos(\frac{\pi}{6}-x)}\ =\ \sin(\frac{\pi}{6})\sin(x)+\cos(\frac{\pi}{6})\cos(x)\)

Now we can see that

\(A\ =\ \sin(\frac{\pi}{6})\ =\ \frac12 \\~\\ B\ =\ \cos(\frac{\pi}{6})\ =\ \frac{\sqrt3}{2}\)

And so...

\(\dfrac{B}{A}\ =\ \dfrac{(\frac{\sqrt3}{2})}{(\frac{1}{2})}\ =\ \sqrt3\)    _

\(f''(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x) \)

Let's plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.

\(f''(0)\ =\ \sin(\frac{0^2}{8})-2\cos(0)\ =\ \sin(0)-2\cos(0)\ =\ 0-2(1)\ =\ -2\ <\ 0\)

Since  f''(0) < 0  the graph of  f  is concave down at  x = 0  and so a local max occurs at  x = 0

\(f''(6.949)\ =\ \sin(\frac{6.949^2}{8})-2\cos(6.949)\ \approx\ -1.817 \ <\ 0\)    (assuming  x  is in radians)

Since  f''(6.949) < 0  the graph of  f  is concave down at  x = 6.949  and so a local max occurs at  x = 6.949

Here's some more info about this:

\(2\cos\theta+\sin(2\theta)=0\)

Let's use the double-angle formula for sin which says  \(\sin(2\theta)=2\sin\theta\cos\theta\)

\(2\cos\theta+2\sin\theta\cos\theta=0\)

Now we can factor  2 cos θ  out of both terms on the left side of the equation.

\(2\cos\theta(1+\sin\theta)=0\)      (Notice if we distribute  2 cos θ  we get the previous expression)

Set each factor equal to  0  and solve for  θ:

\(\begin{array}{ccc} 2\cos\theta=0&\text{or}&1+\sin\theta=0\\~\\ \cos\theta=0&\text{or}&\sin\theta=-1\\~\\ \theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\dots&\text{or}&\theta=\frac{3\pi}{2},\frac{7\pi}{2},\frac{11\pi}{2},\frac{15\pi}{2},\dots\\~\\ \theta=90^\circ,270^\circ,450^\circ,630^\circ,\dots&\text{or}&\theta=270^\circ,630^\circ,990^\circ,1350^\circ\dots \end{array}\)

The solutions in the interval  [0°, 360°)  are:   90°, 270°

\(\sum\limits_{k=1}^{n}(k)(k+1)(k+2)\) .

The question implies that the real part of  \(\frac{1}{1-z}\)  is the same for all nonreal complex values of  z  such that  \(|z| = 1\) . So assuming that's true, we can pick any nonreal complex  z  such that  \(|z| = 1\)  and find the real part of  \(\frac{1}{1-z}\) .

Let's pick  z  =  0 + 1i

Then...

\({\frac{1}{1-z}}\ =\ \frac{1}{1-(0+1i)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\cdot\frac{1+1i}{1+1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-i^2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-(-1)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{2}+\frac{1}{2}i \)

The real part is  \(\boxed{\frac12}\)

We can check a few more cases to see that it is 1/2 for those as well:

Let's find what values of  x  make  f(x) = 25

f(x)  =  25

3x² - 3  =  25

3x²  =  28

x²  =  28/3

x   =  ±√[ 28/3 ]

Now let's plug in both of these into  g(f(x))

g( 25 )   =   g(f( √[ 28/3 ] )   =   ( √[ 28/3 ] )² + √[ 28/3 ] + 1   =   31/3 + √[ 28/3 ]

g( 25 )   =   g(f( -√[ 28/3 ] )   =   ( -√[ 28/3 ] )² - √[ 28/3 ] + 1   =   31/3 - √[ 28/3 ]

And adding these two together we get...

(  31/3 + √[ 28/3 ]  )   +   (  31/3 - √[ 28/3 ]  )   =   31/3 + 31/3   =   62/3