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The main diagonal is 2. Because it is a cube, all the side lengths are the same, and we have the equation: \(\sqrt{x^2+x^2+x^2}=2\)

Solving, we find \(x^2 = {4 \over3}\). The surface area is \(6x^2\), meaning the surace area of the cube is \(8\).

Now, we have the equation: \(4\pi r=8\)

Diving both sides by 4, we get: \(\pi r = 2\)

Solving for r, we find \(r = \color{brown}\boxed{\sqrt{2\over \pi}}\) 

Happy Birthday, Melody !     You don't have to count them any longer, but you have to celebrate them !

 ( What anniversary of your 21st birthday IS it ?!?) 

~ EP

The main diagonal of a cube is \(\sqrt 3\) times longer than its side length.

Therefore, the side length of cube = \(\dfrac2{\sqrt 3}\).

Now, we can calculate the surface area by the formula \(\text{surface area of cube} = 6(\text{side length})^2\).

\(\text{surface area of cube} = 6\left(\dfrac2{\sqrt 3}\right)^2 = 8\)

Since the amount of paint used is exactly the same as the amount of paint to paint the surface of a sphere, the surface area of the sphere is the same as that of the cube.

\(\text{surface area of sphere} = 8\).

Let r be the radius of the sphere. Then,

\(4\pi r^2 = 8\\ r^2 = \dfrac2\pi\\ r = \sqrt{\dfrac2\pi}\)

Therefore, radius of sphere = \(\sqrt{\dfrac{2}\pi}\).

Yes, the polynomial exists:

Happy birthday, Melody!

\(c = \) Number of chicken sandwiches purchased

\(s\) = Number of vegetable soups purchased

\(2c+3s=16.50\)          (1)

\(c+2s=9.50\)              (2)

Subtract \(2s\) from both sides: \(c=9.50-2s\)

Substitute into (1): \(2(9.50-2s)+3s=16.50\)

Simplify: \(19-4s+3s=16.50\)

Solve: \(s = \color{brown}\boxed{2.50}\)

Two cakes are better than just one   !!!!!

Now we have to wait for Melody to do the honors!!

Happy Birthday!!

I hope your day is going well :)

Here is my cake for you!!!!!

We know the number of boys who downloaded the game is\(1,916 \times 6 = 11,496\).

But, only half of these bought the full game, so only \(11496 \div 2 = \color{brown}\boxed{5,748}\) boys downloaded the game. 

Happy Birthday Melody!!!!!!!!

There are \(94 \times 54 = 5076 \) nails. 

He then distributes the \(5076\) nails among \(9\) houses. 

Thus, each house receives \(5076 \div 9 = \color{brown}\boxed{564}\)

I think there is a polynomial that meets these two requirements.

We calculate the area of the triangle.

Area of triangle JKL = \(\dfrac12 \cdot 10 \cdot \sqrt{25^2 - \left(\dfrac{10}2\right)^2} = 50\sqrt 6\)

Then, we use the formula \(\text{circumradius} = \dfrac{\text{product of 3 sides}}{4(\text{area})}\)

\(\text{circumradius} = \dfrac{25(25)(10)}{4(50\sqrt 6)} = \dfrac{125}{24}\sqrt6\)

We visualize the equation as a circle on the coordinate plane.

We first need to find out an explicit form of g(2x).

\(\begin{array}{rcl} g(x) &=& \sqrt{\dfrac{x + 3}{4}} \end{array}\)

Replace x by 2x:

\(\begin{array}{rcl} g(2x) &=& \sqrt{\dfrac{2x+ 3}{4}} \end{array}\)

Now we know what g(2x) and g(x) are, we proceed to solve the equation.

\(\begin{array}{rcl} g(2x) &=& 2g(x)\\ \sqrt{\dfrac{2x + 3}4} &=& 2\sqrt{\dfrac{x + 3}{4}}\\ \dfrac{\sqrt{2x + 3}}2 &=& \sqrt{x + 3}\\ \left(\dfrac{\sqrt{2x + 3}}2\right)^2 &=& \left(\sqrt{x + 3}\right)^2\\ \dfrac{2x + 3}4&=&x + 3\\ 2x+3&=&4(x+3)\\ 2x+3&=&4x+12\\ -9&=&2x\\ x &=& -\dfrac92 \end{array}\)

However, when we substitute \(x =-\dfrac92\) into the equation, we see that \(x = -\dfrac92\) is not in the domain of g(x).

Hence, it is an extraneous root and needs to be rejected.

There is no such value of x that satisfies g(2x) = 2g(x).

Note that \(168|a\) and \(88|a\) by the property of gcd.

Take \(a = \operatorname{lcm}(168, 88) = 1848\).

Then \(b = 168\) and \(c = 88\) satisfies the conditions. You can check that gcd(b, c) is minimum in this case because for any positive integers \(k,l\), \(\gcd(kb, lc) = \gcd(k, l)\gcd(b, c)\gcd\left(\dfrac{k}{\gcd(k,l)}, \dfrac{c}{\gcd(b, c)}\right)\gcd\left(\dfrac{b}{\gcd(b, c)},\dfrac{l}{\gcd(k,l)}\right)\geq \gcd(b, c)\), and b = 168k for some positive integer k and c = 88l for some positive integer l.

\(\min\gcd(b, c)=\gcd(168, 88) = 8\).

Here are the first 6 primes: \(\{2,3,5,7,11,13\}\)

We can take a look at all the possibilities by making a table. 

\(\begin{matrix}&\color{blue}2&\color{blue}3&\color{blue}5&\color{blue}7&\color{blue}11&\color{blue}13\\\color{red}2&4&5&7&\color{magenta}9&13&\color{magenta}15\\\color{red}3&5&\color{magenta}6&8&10&14&16\\\color{red}5&7&8&10&\color{magenta}12&16&\color{magenta}18\\\color{red}7&\color{magenta}9&10&\color{magenta}12&14&\color{magenta}18&20\\\color{red}11&13&14&16&\color{magenta}18&22&\color{magenta}24\\\color{red}13&\color{magenta}15&16&\color{magenta}18&20&\color{magenta}24&26\end{matrix}\)

Blue numbers and red numbers represent the number that Paul chose and the number that Jesse chose respectively.

Pink numbers are those divisible by 3.

By trial-and-error, out of all 36 possible cases, 13 cases has a sum divisible by 3.

Therefore, 

\(\text{Prob}(\text{sum divisible by 3}) = \dfrac{13}{36}\)

Assuming they can pick the same number

36 choices

2 7

2 13

5 7

5 13

7 11

3 3              the red ones can be reversed      11 choices out of 36

OOps...I see from Max answer I missed   11 13      13 11     so the answer would be the same    13 /36          (Thanx, max!)

You can check your answer by substituting \(x = 0\) into the original equation:

\(\begin{array}{rcl} \text{LHS} &=& 400 + 0\\ &=& 400\\ \text{RHS} &=& 200x\\ &=& 200(0)\\ &=& 0\\ &\neq& \text{LHS} \end{array}\)

Then \(x= 0\) does not satisfy the equation, and hence your answer is not correct.

\(\begin{array}{rcl|c} 400+x&=&200x\\ 400+x-x&=&200x-x&\text{subtract }x\\ 400 &=& 199x\\ \dfrac{400}{199} &=& x&\text{divide by }199\\ x &=& \dfrac{400}{199} &\text{exchange sides} \end{array}\)

Note that 

\(\begin{array}{rcl}f(x) &=& (x^2 + 6x + 9)^{50} - 4x + 3\\ &=&\left( (x + 3)^2\right)^{50} - 4x + 3 \\&=&(x+3)^{100}-4x+3\end{array}\)

For each of \(i=1,2,\cdots,100\), since \(x = r_i\) is a root of \(f(x)\):

\(\begin{array}{rcl} f(r_i) &=& 0\\ (r_i+ 3)^{100} &=& 4r_i - 3 \end{array}\)

Now, using the equation above,

\(\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&(4r_1 - 3) + (4r_2 - 3) + \cdots + (4r_{100} - 3)\\ =&4(r_1 + r_2 + \cdots + r_{100}) - 300 \end{array}\)

By Vieta's formula, we have:

\(r_1 + r_2 + \cdots + r_{100} = -\dfrac{\text{coefficient of }x^{99}\text{ in }f(x)}{\text{coefficient of }x^{100}\text{ in }f(x)}\)

Using the binomial theorem, we can expand \((x + 3)^{100}\).

\((x + 3)^{100} -4x+3 = x^{100} + 300x^{99}+\cdots\)

Therefore, 

\(r_1 + r_2 + \cdots + r_{100} = -\dfrac{300}1 = -300\)

Recall that \((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}=4(r_1 + r_2 + \cdots + r_{100}) - 300\). Then:

\(\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&4(-300) -300\\ =& -1500 \end{array}\)

Given:

  2c + 3b = 16.50

  1c + 2b = 9.50       <===== multiply this equation by -2     then add to top equation ( this will eliminate 'c')

  -2c -4b = -19.00     now add the top and the bottom

         -b = -2.50

            b = $ 5.00

Three htirds is one mile    soooo Dennis has run four fourths.