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stop cheating bozo

like 20 or something

idk man solve it for yourself

So this should be an equilateral triangle since two angles are 60 degrees.  This means BD + CD = BC = AB =DC.  The sin(60) is sqrt(3)/2.  So if AD=24, then each side of the equilateral triangle is 16sqrt(3).  Doing the triangle area formula of (BH)/2, we get 192 sqrt(3).  

I got this answer instead using the law of cosine: \[AB^2=25-12\sqrt{3}, AC^2= 20-8\sqrt{3},\] and \[BC^2= 7\]

11, 26, 37, 63, 100

i appreciate your help, but thats incorrect

Let's first find the length of one of the sides of the equilateral triangles. Since each side of the square has length 2, the altitude of one of the equilateral triangles is also 2. Therefore, the length of one of the sides of the equilateral triangle is:

$$s = \frac{2}{\sqrt{3}}$$

Now let's consider one of the smaller squares formed by joining two adjacent vertices of the equilateral triangles. Since the angles of an equilateral triangle are all 60 degrees, the angle between two adjacent sides of the square is 120 degrees. Therefore, the length of one of the sides of the smaller square is:

$$2s \cos 120^\circ = -s$$

The negative sign is because the cosine of 120 degrees is negative.

Now let's consider the diagonal of the large square. This diagonal is the sum of the diagonal of the smaller square and two sides of the equilateral triangles. Since the sides of the equilateral triangles have length $s$, their diagonals have length $2s$. Therefore, the length of the diagonal of the large square is:

$$2(-s) + 2s + 2s = 4s = \frac{8}{\sqrt{3}}$$

The length of a side of the large square is the square root of half the length of its diagonal, so:

$$\text{side length of large square} = \frac{1}{\sqrt{2}} \cdot \frac{8}{\sqrt{3}} = \boxed{\frac{4\sqrt{6}}{3}}$$

The answer is s, v.

The decimal is 13.125

We can do this by using some simple arithmetic: 

\(13 \frac{1}{8} = 13 + \frac{1}{8}\).

This lets us separate the sum and focus on finding \(\frac{1}{8}\). \(\frac{1}{8} = 0.125\) from simple long division, so we have 

\(13 + \frac{1}{8} = 13+ 0.125 = 13.125\). 

Guest, I assume you mean how many ways you can put 7 textbooks on the bookshelf . 

If this is true, we can use some simple counting to solve this. We know that this bookshelf has to be of the form

_ _ _ M _ _ _ 

if M is a math textbook. So we simply need to find ways to arrange the other textbooks. We know all of the textbooks are different from the problem, so we do not have to worry about distinguishability. Thus, there are \(6! = 720\) ways of arrangement. 

Because 493 is neither a perfect square nor is divisible by any perfect squares, it cannot be simplified further from \(\sqrt{493}\). The best we can do is use some rounding to get 

\(\sqrt{493} \approx 22.203603...\)

Alternative Solution: 

Notice that \((x+y)(x-y) = x^2-y^2\), so simply multiplying the two together gives: 

\((x+y)(x-y) = x^2-y^2 = \frac{15}{7} \cdot \frac{7}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}\)

\({1 \over \sqrt{2} + \sqrt{8} + \sqrt{32} }\)

\({1 \over \sqrt{2} + 2\sqrt{2} + 4\sqrt{2} }\)

\({1 \over 7\sqrt{2} }\)

\({1 \over 7\sqrt 2} \times { \sqrt{2} \over \sqrt {2}} \)

\({\sqrt 2 \over 14} \)

\(2 + 14 = \color{brown}\boxed{16}\)

The denominator of 1/(sqrt(2) + sqrt(8) + sqrt(32)) can be rationalized by multiplying both numerator and denominator by -sqrt(2) + sqrt(8) + sqrt(32). This gives us (-sqrt(2) + sqrt(8) + sqrt(32))/(-2 + 8 + 32) = sqrt(2)/19.

The final answer is 2 + 19 = 21.

The tenth term of the sequence an can be found by substituting n = 10 into the given recurrence relation and iteratively solving for the value of a10. The result is a_10 = 1/2.

If 2 freshmen are at both ends, there are \(5 \times 4 \times 3 \times 2 = 60\) ways to order the freshman. 

Then, there are \(5!\) ways to order the rest (note that there will always be 2 sophomores next to each other). 

So, there are \(5! \times 60 = \color{brown}\boxed{7200}\) ways to order the students. 

Adding the two equations gives us \(2x = {184 \over 63}\), meaning \(x = {92 \over 63}\).

This means that \(y = {43 \over 63}\), so \(x^2 - y^2 = ({92 \over 63})^2 - ({43 \over 63})^2 = \color{brown}\boxed{5 \over 3}\)

We need to split this problem up into 4 cases: 

Case 1 - The last number is not a 1, 2, or 4: There are \({7! \over 3! 2!} = 420\) cases

Case 2 - The last number is a 1: There are \({7! \over 3!2!2!} = 210\) cases

Case 3 - The last number is a 2: There are \({7! \over 3!3!} = 140\) cases

Case 4 - The last number is a 4: There are \({7! \over 4!2!} = 105\) cases

So, there are \(420 + 210 + 140 + 105 = \color{brown}\boxed{875}\) cases.

Continuing with Melody's excellent hint, if you need help solving it, here it is:

\(\begin{align*} x^2+8x-884&=0\\ x&=\dfrac{-8\pm\sqrt{64+3536}}{2}\\ x&=\dfrac{-8\pm\sqrt{3600}}{2}\\ x&=\dfrac{-8\pm60}{2}\\ x&=-4\pm30\\ x&=-34, 26 \end{align*}\)

Obviously, the answer cannot be -34 (imagine -34 rows). So the answer is 26.

This is how I would look at it.

Choose 2 dogs from  12 long haired ones   and    Choose 3 cats from ll long haireds ones 

and put it over  5 animals from a total of 30 animals

= ( 12C2 * 11C3 ) / 30C5    =   66*165 / 142506  =  10890 / 142506 = 605 / 7917   = approx 7.6%

That sounds ok to me.

I do not know why you multiplied that by 5C2