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To get g(8), we must force f(x) = 8 to happen.

When f(x) = 8, 

\(x^2 - 6x - 4 = 0\\ x = 3 \pm \sqrt{13}\)

Then g(8) is either 2(3 + sqrt(13)) + 3 or 2(3 - sqrt(13)) + 3. Now it is your turn to simplify these expressions.

\(2x^2 - 6x + 8 = 2(x^2 - 3x) + 8=2\left(x^2 - 3x + \dfrac94\right) +\dfrac72 = 2\left(x - \dfrac32\right)^2 + \dfrac72\)

Now you should be able to determine the values of a, b, and c.

\(z^2 + z + 1 = 0 \\ (z^2 + z + 1)(z - 1) = 0\\ z^3 - 1 = 0\\ z^3 = 1\)

So, if z^2 + z + 1 = 0, then z^3 = 1. (Note that the reverse is not true: z^3 = 1 does not imply z^2 + z + 1 = 0.)

\(\quad z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\\ = z^{48}(z + z^2 + z^3 + z^4 + z^5)\\ = (z^3)^{16} (z + z^2 + z^3(1 + z + z^2))\\ = 1^{16} (z + z^2 + z^3 (0))\\ = z + z^2 \\ = (z^2 + z + 1) - 1\\ = 0 - 1\\ = -1\)

Just move the 1 to the right-hand side and it's done.

\(z^2 + z + 1 = 0\\ (z^2 + z + 1) - 1 = 0 -1\\ z^2 + z = -1\)

3000 / 500 ==6 inches on the map.

15 / 25 ==0.6  x  100 ==60%

15 is 60% of 25

z^2 + z = ((-1+-isqrt(3)))^2/4 + (-1+-isqrt(3))/2 Then just add them up by multiplying the right one by 2/2

if \(z^2 + z + 1 = 0\),
find \(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\).

\(\begin{array}{|lrcll|} \hline & z^2 + z + 1 &=& 0 \\ \Rightarrow & z + 1 &=& -z^2 \\ \Rightarrow & z^2 + z &=& -1 \\ \hline \end{array} \begin{array}{|rcll|} \hline z^2 + z + 1 &=& (1+z)^2-z \quad | \quad z + 1 = -z^2 \\ z^2 + z + 1 &=& (-z^2)^2-z \\ z^2 + z + 1 &=& z^4-z \quad | \quad z^2 + z + 1 = 0 \\ 0 &=& z^4-z \\ \mathbf{z^4} &=& \mathbf{z} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{49} &=& z^{48}z \\ z^{49} &=& \left(z^{4}\right)^{12}z \quad | \quad z^4 = z \\ z^{49} &=& \left(z\right)^{12}z \\ z^{49} &=& \left(z^{4}\right)^{3}z \quad | \quad z^4 = z \\ z^{49} &=& \left(z\right)^{3}z \\ z^{49} &=& z^{4} \\ \mathbf{z^{49}} &=& \mathbf{z} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && z^{49} + z^{50} + z^{51} + z^{52} + z^{53} \\ &=& z^{49}\left(1 + z + z^2+ z^3 + z^4\right) \\ && \boxed{ z^{49}=z \\ 1 + z + z^2 = 0 \\z^4 = z } \\ &=& z\left(0+ z^3 + z\right) \\ &=& z^4+z^2 \quad | \quad z^4 = z \\ &=& z+z^2 \quad | \quad z^2 + z = -1 \\ &=& -1 \\ \hline \end{array}\)

\(\boxed{z^{49} + z^{50} + z^{51} + z^{52} + z^{53} = -1}\)

https://web2.0calc.com/questions/sum_52#r2

The expression can be factored by using the X method or the multiplying method,

which is (3x+4)(x+2). A and B are 4 and 2, so A - B = 2. btw the multiplying method is where you multiply 8 by 3 and get 24, and 12*2=24 12+2=14,

3*4=12, so (3x+4) comes together. then the 2 is prime, so it goes on the other side. ( it's a little confusing at first but it gets better )

What constant will make y^2 + 28y + 16 + k the square of a binomial?  

This one is easy because the coefficient of the squared term is 1. 

Remember that in general terms (x + a)²  =  x² + 2ax + a²   

So, for y² + 28y + a²    the value represented by 2a is 28   

You just take half of 28 to get "a"      a = 28/2 = 14   

Of course, you want a² so square 14 to get 196  

But you already had 16 so k = 196 – 16    thus    k = 180  

_.

A company sells bath oils contained in spherical, water-soluble pods. The circumference of each spherical pod is 37.68 millimeters. What is the volume of a pod? (Use 3.14 for . Round the answer to the nearest tenth, if necessary. Recall that the formula for the volume for a sphere is [missing in the original] and the formula for the circumference of the great circle is [missing in the original].)  

The circumference of a circle is 2 π r    so    (2)(3.14) r = 37.68  

         r = 37.68 / 6.28   

         r = 6  

The volume of a sphere is (4/3) π r³   

         V = (4/3)(3.14)(6³)   

         V = (4/3)(3.14)(216) = 904.32     round to nearest tenth per problem = 904.3 mm³  

_.

Help pls

\(\text{In triangle $ABC$, points $D$ and $F$ are on $\overline{AB}$,$\\$ and $E$ is on $\overline{AC}$ such that $\overline{DE}\parallel \overline{BC}$ and $\overline{EF}\parallel \overline{CD}$.$\\$ If $AF = 9$ and $DF = 3$, then what is $BD$?} \)

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{\overline{AC}}{\overline{AE}} &=& \dfrac{9+3}{9} \\\\ (2) & \dfrac{\overline{AC}}{\overline{AE}} &=& \dfrac{9+3+BD}{9+3} \\ \hline \\ (1)=(2): &\dfrac{\overline{AC}}{\overline{AE}}= \dfrac{9+3}{9} &=& \dfrac{9+3+BD}{9+3} \\\\ & \dfrac{9+3}{9} &=& \dfrac{9+3+BD}{9+3} \\\\ & \dfrac{12}{9} &=& \dfrac{12+BD}{12} \\\\ & \dfrac{4}{3} &=& 1+\dfrac{BD}{12} \\\\ & \dfrac{BD}{12}&=& \dfrac{4}{3}-1 \\\\ & \dfrac{BD}{12}&=& \dfrac{1}{3}\\\\ & BD &=& \dfrac{12}{3} \\\\ & \mathbf{BD} &=& \mathbf{4} \\ \hline \end{array}\)

Thank you so much! I didn't think of subtracting 1!

Why?

Hi Atlas,

I don't know the answer, I am just playing with the question.

We have     \(t_n=\frac{n(n+1)}{2}\)

so              \(4t_n=2n(n+1)\)

the questions says what is the greatest possible value of the GCD of     2n(n+1)    and   n-1   

If n=3 then the greatest divisor is 2

If n=5 then the greatest divisor is 4

I don't know what the biggest one is though.  Maybe 4 ?????

What is sin RPS ?

sin RPS = sin (180° - RPQ) = 7/20

In order to find k, you need to plug (\(\frac{1}{4}\),-6) into the equation and solve for k.

-\(\frac{1}{2}\) - kx = 6y

Plug in x and y (x,y)

-\(\frac{1}{2}\) - k(\(\frac{1}{4}\)) = 6(-6)

Simplify

-\(\frac{1}{2}\) - \(\frac{1}{4}\)k = -36

Add \(\frac{1}{2}\) to both sides

In order to do this, -36 turns into -\(\frac{72}{2}\) 

-\(\frac{1}{4}\)k = -\(\frac{71}{2}\)

Divide each side by -\(\frac{1}{4}\)

This means it's actually being multiplied by -4: -\(\frac{71}{2}\)(-4)

k = \(\frac{284}{2}\)

Simplify

k = 142

Alternatively:

-\(\frac{1}{4}\)k = -35.5 instead of -\(\frac{71}{2}\)

Multiply by -4 on both sides

k = 142

It saves you the step of simplification, but some prefer to keep the fraction instead of switching to decimals. Figured I would show both!

I thought the denominator in the agm formula needed to be the number of variables? 

For the first one, you get the minimum when x = y = z.  So the minimum value is (x + 2x + 4x)/(4/x + 2/x + 1/x) = 49.

For the second one, you get the minimum when x = y = z.  So the minimum value is (x^3 + 5x^3 + 25x^3).(x^3) = 31.

Find the value of:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\) ?

\(\begin{array}{|rcll|} \hline x &=& 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \ddots}} \\ && \boxed{x-1= \cfrac{1}{2 + \ddots} } \\ x-1 &=& \cfrac{1}{2 + (x-1) } \\ x-1 &=& \cfrac{1}{x+1} \\ (x-1)(x+1) &=& 1 \\ x^2-1 &=& 1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

If the perimeter is 12 one side can't be 20 leathcheann.

Deleted because it was incorrect. Guest's answer is correct.