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What is the numerical value of the product of the lengths of all diagonals of a regular pentagon of side length 1?

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\(\alpha=\frac{180°-72°}{2}=54°\\ sin\ \alpha =\frac{d}{2\cdot 1}\\ d=2\ sin\ \alpha=2\ sin\ 54°\)

\(d=1,618\)

The numerical value of the product of the lengths of all diagonals is \(1.618^5=11.09\)

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Where is F in the Pentagon ABCDE?

iirc, this is very similar to a previous mathcounts nationals problem. here is the solution to that from my document

CFD is a 90, 45, 45 triangle, so CF and DF is $\sqrt2$
$BE = \sqrt(2(2+\sqrt(2))^2) = \sqrt{2} \cdot (2 + \sqrt{2})= 2\sqrt{2} + 2$
$AE^2 = AB^2 + BE^2 = 2 \times (2\sqrt{2} + 2)^2$
$AE = \sqrt{2} \times (2\sqrt{2} + 2) = 4 + 2\sqrt{2}$
$a + b + c = 2 + 4 + 2= \boxed{8}$

You are essentially choosing three items from six

$\dbinom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = \boxed{20}$

Two of the exterior angles of a triangle are 158 degrees and 93 degrees Find the third exterior angle, in degrees.     

The easy way: 

By knowing that the sum of the exterior angles of any convex polygon is 360 degrees.  

Therefore, the exterior angle at the third vertex can be found as 360^o – (158^o + 93^o)  =  109^o  

The lengthy way:  

Call the triangle vertices A, B, and C.  Extend one side of the triangle at each vertex. 

A straight line is 180^o so find the supplementary angle at each vertex to determine the size of the third angle. 

Call A the vertex with exterior angle 158^o  so the interior angle at vertex A is 180^o – 158^o = 22^o.  

Call B the vertex with exterior angle   93^o  so the interior angle at vertex B is 180^o –   93^o = 87^o.  

So the interior angles of the triangle are C^o and 22^o and 87^o.  

The sum of the interior angles of a triangle is 180^o so angle C is 180^o – (22^o + 87^o)  =  71^o  

The exterior angle at C is the supplementary angle of 71^o which is 180^o – 71^o  =  109^o   same as above  

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Ok, so I think the answer is the square root of 70. Based on the information I drew out a picture. Then I made two equations: 5^2 + x^2 = a^2 and 14^2 + x^2 = b^2. Because triangles ABD and CBD are right triangles I used the Pythagorean theorem to create those two equations. Then I can add a^2 and b^2 together to get c^2. So, I add both equations together: 5^2+x^2 + x^2 +14^2 = 19^2. Then solving the equation: 5^2 + 14^2 + 2x^2 = 19^2. I end up with x^2 = 70 which is the square root of 70. I hope this helps! 

lets say n = 10

101  / 7 = 14 R 3

To calculate

the number of sports cards Miguel had.

Let y be the initial Corey's sports cards and x be the initial Miguel's sports cards then according to the question we have,

⇒ Initial Corey's sports cards= (\(1 {3 \over 4}{}{}\)) (Initial Miguel's sports cards)

⇒ y = (\( {7 \over 4}{}{}\))(x)

⇒ y =\({7x \over 4}{}{}\)

After giving half of Miguel's sports cards to his friend, Miguel's final sports cards are

\(x\)-\( {x \over 2}{}{}\) and Corey's sports card are \(y\)+\( {x \over 2}{}{}\).

According to the question,

⇒ Final Miguel's sports cards=(Final Corey's sports cards)−18

⇒ (x - \({x\over 2}{}{}\)) = (y + \( {x \over 2}{}{}\)) - 18

⇒ \(y\) = 18

Plug it above,

⇒ 18 = \( {7x \over 4}{}{}\)

⇒ \(x\) = \({18 * 4\over 7}{}{}\)

⇒ \(x\) = 10.29

⇒ \(x\) = 10

Thus, Miguel had 10 sports cards at first.

BI = 1 + sqrt(2).

The radius is 5*sqrt(2).

Angle A = 30 degrees.

Tell whether the difference between the two integers is always, sometimes, or never positive. A negative integer and a positive integer     

Sometimes.  You can tell by trying examples; all you need is one of each.  

The tricky one is when you subtract a negative number, like a – (–b). 

The negative, times a negative, makes a plus, you end up with a + b. 

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This is , $(x - 4)(2y + 1) = 1 + xy.$

Solving, give $x = 9, y = 6.$ So, the product is $9 \cdot 6 = \boxed{54}$.

The graph is of $f(x).$ $f(2)$ is equal to the $y$ value when $x = 2,$ so $f(2) = \boxed{-2.}$

$5^8 + 5^7 + 5^6 + 5^5 + 5^4 + 5^3 + 5^2 + 5^1 = \boxed{488280}$ words, as there can be words with 1-7 letters as well.

First auto has a   2 x 60 = 120 mile headstart

Closing speed of the second auto is  80 - 60 = 20 m/hr      it will take    120 m / 20 m /hr = 6 hours to catch the first auto

180-x = 16 (90-x)

180 - x = 1440 -16x

15x = 1440 - 180

x = 1260/15 = 84   degrees

5 choices for each position in a word

5 * 5 * 5 * 5 * 5 * 5 * 5 * 5 =

5⁸    words

[10^18 - 1] / 9 ==111,111,111,111,111,111 [18-digit number]

111111111111111111 mod 9 ==0

111111111111111111 mod 21==0

q = 1.5 p    =  q/1.5 = p

r = 1.5 q      Sub in the red equation

r =  1.5 (1.5 p)

r/p = 2.25

Since repeats are allowed, then there are:

100 x 100 ==10,000 2-digit products.

It turns out that there are: 3,600 ways of getting a multiple of 5 out the total of 10,000 outcomes.

Original measure    a / 360 * pi * d               a = arc measurement in degrees    d = diameter

now increase      a * 1.20   / 360   * pi * d * 1.35

ratio of increase to original

1.2 * 1.35 / 1 * 1  =   1.62             this is a 62 % increase

133 ones   [ 4 dice beginning with 1]
152 twos  [4 dice beginning with 2 ]
189 threes[4 dice beginning with 3]
152 fours  [4 dice beginning with 4]
133 fives  [4 dice beginning with 5 ]
216 sixes [4 dice beginning with 6 ]
Total number of all possible permutations of the products of 4 dice that are all even and divisible by 3== 133 + 152 + 189 + 152 + 133 + 216==975
The probability is: 975 / 6^4==325 / 432

What is the combined area of the two gray regions?

The combined area of the two gray regions is \(20-2\cdot\frac{1}{6}\cdot10=\color{blue}16\frac{2}{3}\)

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