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small top      4x20  cm²

large top      10x20

bottom         14 x20

back             (8+2) x 20

Front            (2+8) x 20

Sides (two)           2   *    [  2x14    + (8x4) ]

Add 'em all together......

The axis of symmetry is the 'x' value of the vertex, which can be found by   x = -b / 2a    where a = -5   b=30

    or you can find the two zeroes of the equation.....the axis of summetry is exactly between the zeroes.

A teenager is someone who is 13, 14, 15, 16, 17, 18 or 19 years old. The product of the ages of a particular group of teenagers is 705,600. What is the mean of their ages in years? Express your answer as a decimal to the nearest tenth.  

I could find only one way to produce a product of 705,600 by

using only "teen" numbers – namely, 14 • 14 • 15 • 15 • 16 

so there are five teenagers in the group, with those ages. 

                                                                             74   

The mean (average) of those five numbers is     ——     =     14.8 

                                                                              5   

_.

Let X be total number of membersX = r m (1)X = (m + 1) *(r -2) = r m + r – 2m – (2)(2) – (1): r – 2m – 4 = 0  r = 2m (3)(3) to (2): X = (m+1)(2m+4-2)=2(m+1) (4)Since X < 100, (m+1)2 = X/2 < 50Largest possible integer m is 7. Answer: X = 98
 

For a triangle exists, the side lengths of the triangle must be such that the sum of the two shorter sides must be greater than the third side.
This also is equivalent to any two sides must have a sum greater than the third side.
So
7+10 > x^2, => x^2 < 17 => x < sqrt(17)     (maximum)
7+x^2 > 10, => x^2 >3 => x > sqrt(3)
Therefore
sqrt(3) < x < sqrt(17)
If x must be an integer,
2< x < 4, or x = {2,3,4}

I'm going to assume you mean the angle to be \(\frac{\pi}{5}\).

You have an error here:::

EF ≠ 1/2 * BE                    EF = 2/3 * BE

The centroid divides a median into two line segments, and their ratio is 2 : 1

decagon rotation,

Are you sure you don't mean \(\sin^2(\frac{\pi}{5})\) ?

I get 72 degrees,

I suppose I check it with a GeoGebra diagram.

Let line AC be perpendicular to line CE. Connect A to the mid-point of CE which is D, and connect E to the mid-point of AC which is B.  If AD and EB intersect at point F, BC=CD=15cm, then find the area of △DFE.

AD = BE = sqrt( 30² + 15² )

FD = 1/3AD

FE = 2/3BE

Now we have all 3 sides of a triangle DFE.

We can use Heron's formula to calculate the area of this triangle.

Area of  △DFE = 75 cm²  

Why does your decagon have 11 vertices?     A.....K

3x-5+2x = 90

5x=95

x=19

Hahahaha... it's the same as 5-gon's interior angle.

Let \(x\), \(y\), and \(z\) be positive real numbers such that  \(xyz=1\)
Find the minimum value of \((x+2y)(y+2z)(xz+1)\).

\(\mathbf{\huge{AM \geq GM }}\)

\(\begin{array}{|rcll|} \hline \dfrac{x+2y}{2} & \ge & \sqrt{x*2y} \\\\ \dfrac{y+2z}{2} & \ge & \sqrt{y*2z} \\\\ \dfrac{xz+1}{2} & \ge & \sqrt{xz*1} \\\\ \hline \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y}\sqrt{y*2z} \sqrt{xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y*y*2z*xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{4x^2y^2z^2} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & 2xyz \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*xyz \quad | \quad xyz=1 \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*1 \\\\ \mathbf{(x+2y)(y+2z)(xz+1)} & \ge & \mathbf{16} \\ \hline \end{array}\)

The minimum value of \((x+2y)(y+2z)(xz+1)\) is \(\mathbf{16}\)

2. Using the determinant and Cramer's Ruler, we get that k = 28.

3. 

productfor(n, 1, 2004, (n^2 +2*n-1)/(n^2 +n+2#2 - 2) = 831.083979 

The probability is (1/4*1/3*1/2)^4*4! = 1/13824.

f(5*pi/6) + f(5*pi/4) + f(11*pi/6) = 2 - sqrt(3).

The product telescopes to 1002*sqrt(2) - 586.

If the problem is:  -5 < x   then  x > -5    --->    (-5, infinity)

Use parentheses because the -5 is not included; infinity always gets a parenthesis.

i think this is wrong....

or right...

using Lagrange multipliers you can show that the minimum occurs when a=b=c=d=1/4 and is equal to 4/3

but you can also argue this from symmetry since they are all required to be positive.

By the quadratic formula, the roots are equal to

−p±p2−4q−−−−−−√2−p±p2−4q2

So, since the square root and pp are both integers (since otherwise the roots are not rational). But these two integers are either both even or both odd (why?), and therefore their sum or difference is always even - hence your claim.

(5C1+5C2+5C3) * (6C2+6C3+6C4)