NotThatSmart

We can complete this problem in two different ways. 

The first tactic is to essentially compare this root to the quadratic equation of \(5x^2+21x+v = 0\)

From this quadratic, we can identify that a = 5, b = 21, c = v. 

We have the equation

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\\ \frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\\ \sqrt{201} = \sqrt{441-20v}\\ 201 = 441-20v\\ v=12\)

This is a bit complicated and takes a lot of computations, but it does give us the correct answer. 

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The second tactic is to use conjugations of square roots. 

This is because the conjugate root theorem states that if a root of a polynomial is a square root \(a+\sqrt b\), then its conjugate, \(a-\sqrt b\) is also a root

We can apply that to this problem. If  \(\frac{-21-\sqrt{201}}{10}\) is a root, then \(\frac{-21 + \sqrt {201}}{ 10 }\) is also a root. 

The product of the roots is \([ (-21)^2 - 201 ] / 100 = 240/100 = 2.4\)

However, in the quadratic, we also have that \(v / 5\)

Thus, we have 

\(v/5 = 2.4\\ v = 12\)

SO 12 is the final answer. 

Thanks! :)

You are missing some values in your question, but I'm gonna assume the following was asked. 

A machine randomly generates one of the nine numbers   1,2,3...8,9   with equal likelihood. What is the probability that when Tsuni uses this machine to generate four numbers their product is divisible by 4. Express your answer as a common fraction.

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Now, since we can draw two of the same numbers and order doesn't matter, we don't do \(9\choose 4\) to find the total amount of possibilites, but instead just do

\(9 \cdot 9\cdot 9 \cdot 9 = 9^4 = 6561\)

Now, we can use complementary counting to find the number of cases that DO NOT work and then subtract it from 1. 

This is because it's MUCH easier to calculate odd numbers rather than even numbers. 

Our first case is that all 4 numbers are odd. If all 4 numbers are odd, the product is odd, therefore meaning it cannot possibly be divisble by 4. 

Since out of the 9 numbers, 5 numbers are odd, we have a \(5/9\) chance each generation that the number is odd. 

Since we need 4 odd numbers in a row, we simply have \(({5 \over 9})^4 = {625 \over 6561}\)

Our next case is where we get 3 odd numbers in a row and then a 2 or a 6. If we roll a 4 or 8, the product will automatically be divisble by 4. 

Thus, we have \({2 \times 5^3 \times {4 \choose 3} \over 6561} = {1000 \over 6561}\)

Subtracting the sum from 1, we get the final answer of \(1 - ({625 \over 6561} + {1000 \over 6561} ) = \color{brown}\boxed{4936 \over 6561}\)

I would like verification of the final answer, but I believe this is the correct answer. 

Thanks! :)

First, we have to simplify the left side of the equation. 

Expanding it out, we get

\(2x^2-x-28=-31+jx\)

Moving all terms to one side and combining all like terms, we achieve

\(2x^2-x-jx-28+31=0\\ 2x^2-\left(j+1\right)x+3=0\)

If there's only one solution, then that must mean that the descriminant is 0. 

The descriminant of this quadratic is 

\((j+1)^2 - 4(2)(3)\\ j^2+2j-23\)

Setting this to 0, we can now solve for j. We get

\(j^2+2j-23=0\)

Unfortunately, j cannot be factored conveniently, so we must use the quadratic equation. (I'm going to leave that for you). 

From that, we find that

\(j=-1+2\sqrt{6},\:j=-1-2\sqrt{6}\)

I'm pretty sure this is the correct answer, but you can check and verify. 

Thanks! :)

mmm...take a look at the points given. 

\(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\)

This is essentially just the perimeter of the decagon \(P_1P_2P_3P_4P_5...P_{10}\)

The lenngth of one side is essentially

\(radius/2 ( -1 +\sqrt 5)\)

Now that we know one side, we can find the perimeter through the equation

\(10 (1/2) ( -1 + \sqrt 5) = 5 ( -1 + \sqrt 5) ≈ 6.18\)

Thus, the answer is 6.18, 

Thanks! :)

Let's set up a system of equations to solve this problem. 

Let's let b be bagels

Let's let t be toast

And let's let m be muffins. 

From the problem, we have the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

Now, we want everything to be in terms of bagels. 

From the first two equations, we get the equations

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Now, plugging these values into the third equation, we get

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is 262 cents or about 261 and 2/3 cents. 

Thanks! :)