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Here are a couple of possibilities:

Let  a and b be real numbers.
One of the roots of \(x^3 + ax + b = 0\) is \(1 + i \sqrt{3}\).  
Find \(a+b\).

\(\text{Let $x_1=1+i \sqrt{3}$ } \\ \text{Let $x_2=1-i \sqrt{3}$ } \)

\(\begin{array}{|rclrcl|} \hline &&\mathbf{ x^3+ax+b } \\ &=& (x-x_1)(x-x_2)(x-x_3) \\ &=&\left(x-(1+i \sqrt{3})\right) \left(x-(1-i \sqrt{3}) \right)(x-x_3) \\ &=&\left((x-1)-i \sqrt{3}\right) \left((x-1)+i \sqrt{3} \right)(x-x_3) \\ &=&\left((x-1)^2-(i \sqrt{3})^2\right) (x-x_3) \\ &=&\left(x^2-2x+1+3\right) (x-x_3) \\ &=& (x^2-2x+4 ) (x-x_3) \\ &=& x^3-x^2x_3-2x^2+2xx_3+4x-4x_3 \\ &=& x^3-\underbrace{(x_3+2)}_{=0}x^2+\underbrace{(2x_3+4)}_{=a}x\underbrace{-4x_3}_{=b} \\\\ && \begin{array}{|rclrcl|} \hline x_3+2 &=& 0 \\ \mathbf{x_3} &\mathbf{=} & \mathbf{-2} \\\\ a &=& 2x_3+4 \\ &=& 2\cdot(-2) + 4 \\ \mathbf{a} &\mathbf{=} & \mathbf{0} \\\\ b &=& -4x_3 \\ &=& -4\cdot(-2) \\ \mathbf{b} &\mathbf{=} & \mathbf{8} \\ \hline \end{array}\\ \hline \end{array}\)

\(a+b = 0+8 = 8\)

thanks!

\(f(x) = a x^3 + b x^2 + c x + d\\ f(-1) = -a + b - c + d = 15\\ f(0) = d = 0\\ f(1) = a + b + c + d = -5\\ f(2) = 8a+4b+2c+d=12\\ \text{I like to use matrices, there are other ways to solve the following}\\ \begin{pmatrix} -1&1 &-1 \\ 1 &1 &1 \\ 8 &4 &2 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}= \begin{pmatrix}15\\-5\\12\end{pmatrix}\)

\(\text{Gaussian reducing this we get}\\ \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}2\\5\\-12\end{pmatrix}\)

\(f(x) = 2x^3 + 5x^2 -12x = \\ x(2x^2 + 5x - 12) = \\ x(2x-3 )(x+4 )\\ \text{x intercepts at }x \in \left\{0,~\dfrac 3 2,~-4\right\}\)

Hallo CPhill,

Thank you,

the answer

i found it myself.

thanks for the help #3 should've been yes but overall they're correct 

Draw the octagon with a triangle stuck to one of the sides.

The sum of all the angles in a polygon with x sides is (x -2)(180^o)

The interior angles of the octagon are (8 - 2)(180^o) = 1080^o

So one angle of the octagon is 1080^o/8 = 135^o

Add that to the 60^o of the triangle angle touching the octagon angle

and you get a total of 195^o which makes the angle on the other side

which is an interior angle of our n-gon be 360^o - 195^o = 165^o 

I'm going to drop the degrees sign for convenience

Since n is the number of sides of our n-gon,

and there are the same number of angles as there are sides,

the total of the interior angles of our n-gon is (165)(n)  

But the total is also (n - 2)(180) so set them equal

165n = 180n - 360

165n -180n = -360

multiply both sides by -1

180n -165n = 360

15n = 360

n = 24

I'd say 1 and 4 are the yes answers

2 and 3 should be no

\(h(y) = \dfrac{1+y}{2-y}\\ x = \dfrac{1+y}{2-y}\\ (2-y)x = 1+y\\ 2x-1=(1+x)y\\ y = \dfrac{2x-1}{x+1}\\ \text{now substitute }x \to y\\ h^{-1}(y) = \dfrac{2y-1}{y+1}\\ h^{-1}(5) = \dfrac{2(5)-1}{5+1}=\dfrac{9}{6}=\dfrac 3 2\)

\(\sum \limits_{k=2}^{100}\dfrac{1}{\log_k(100!)} = \\ \sum \limits_{k=2}^{100}\dfrac{\log(k)}{\log(100!)} = \\ \sum \limits_{k=2}^{100}\dfrac{\log(k)}{\sum \limits_{j=2}^{100}\log(j)} =\\ \dfrac{\sum \limits_{k=2}^{100} \log(k)}{\sum \limits_{j=2}^{100} \log(j)} = \\ 1\)

Wait who sent your so bored playing Minecraft because I swear Kabob this is the real Noah by the way

that I did not send this

The "distance" can be represented by a segment drawn from the center that is a perpendicular bisector of the chord......call the point of intersection, E......call the center of the circle O....so OE  = 2

So...we have a right triangle  with  the radius = the hypotenuse = 3....segment OE  being a leg = 2   and ED  being the other leg

So 1/2  of the chord length  =  sqrt [  radius^2 - 2^2 ]  =  sqrt [ 3^2 - 2^2]  =  sqrt [ 9 - 4 ]  = sqrt (5)

And chord  CD is twice this length  =  2sqrt (5)

That looks great to me! In general, the formula for a hexagon is \(\frac{3\sqrt{3}}{2}s^2\) . 

Well...I'm not great at Geometry.....but....I can do a little   !!!!

Not too difficult.....the radius of the circle, r   = the side of the hexagon, s...so

324 pi  = pi* r^2 

324  = r^2

18  = r

The hexagon =  six equilateral triangles  ....its area  =

6 * √3 /4  * s^2   =

(3/2)√3 * 324  =

486√3    units^2

OMGosh that is amazing and this is one of the reasons I come here when I need help!

OK....I think I have it.....letting BC = 15  and  and BA    = 20

Then AC  = 25

And the product of the legs BC * BA  / 2  =  the area  = 150

The circle is centered at  (0, 7.5)  and has a radius of 7.5

Notice in the diagram....angle ADB  = 90°.....so  BD  is an altitude of triangle ABC

So...letting AC  be a base...we have that

Area of ABC  = (1/2) base * altitude

150 = (1/2)AC * ( BD)

150  = (1/2)AC * BD

150  = (1/2)(25) (BD)

300 = 25 (BD)

300 /25  = BD  =   12