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First, we should find out the formula for the area of an octagon (if you already know this Guest, just scroll down).

This is the formula:

\(A=2(1+\sqrt2)a ^2\)

(where a is the length of a side)

Since we know that the length is 12, we can easily replace the a for 12.

\(A=2(1+\sqrt2)12 ^2\)

If you are lazy like me, you will dutifully plug it into a calculator and achieve the approximate result of 

\(695.29 \) \(cm ^2\)

The value rounded to five decimal places is 

\(695.29351\) \(cm ^2\)

Hope this helped!

The smallest distance is 5*sqrt(7).

In an equilateral triangle, the circumcenter and incenter coincide, and the radius of the circumcircle is equal to the length of the side of the triangle, while the radius of the incircle is equal to a/2√3.

The area of the circumcircle is given by πr^2, where r is the radius of the circumcircle, so x = πa^2/4.

The area of the incircle is given by πr^2, where r is the radius of the incircle, so y = πa^2/12√3.

Therefore, x - y = πa^2/4 - πa^2/12√3 = πa^2/4 - πa^2/4sqrt(3)

Hence, x - y = πa^2/4 - πa^2/4sqrt(3)

x^2 + y^2 = 54^2 is a circle with radius 54

y= -71/66*x + 71 is the line of travel

find where the line intersects the circle

then find the distance between those 2 intersection points

the points are (53.4, 8.03) and (18.22, 50.83)

distance between them = about 62.23 miles within the radio transmitter

By the Pythagorean Theorem, the distance from Braden from his grandparents' house is sqrt(15^2 + 17^2) = sqrt(514).

The shortest altitude is 40.

The smallest b that works is b = 23: 2*23 + 3 = 49, which is a perfect square.  Isn't that interesting!

How can the 100th term be 97 when at least half a dozen perfect squres, perfect qubes and perfect powers are excluded between 1 and 100 ?! Which means that the 100th term is greater than 100.

To the OP: Don't be lazy! List all the numbers between 1 and 100 and remove all perfect squares, perfect cubes and perfect powers and count what it remains. If you don't know what a perfect square or perfect cube or perfect power is, try this on your calculator: 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25......10^2=100 and so on. 1, 4, 9, 16, 25......100 are perfect squares.

Do the same thing with perfect cubes: 1^3 =1,  2^3=8, 3^3=27, 4^3=64.......and so on. 1, 8, 27, 64.....are perfect cubes.

For perfect powers, generally all perfect squares and perfect cubes are perfect powers as well. However: 2^3=8, 2^4=16, 2^5 =32, 2^6=64........etc. Then: 3^1=3, 3^2=9, 3^3=27, 3^4=81....and then try 4^1=4, 4^2=16, 4^3=64......and so on.....Then try 5, 6, 7, 8, 9, 10......etc. 

The greatest height is 30 ft.

0.80==1 * (1/2)^(100/h), solve for h

h==310.628 - days - half-life of this substance.

UV = 50/17

Thank you so much for this answer and all the hard work you put into it!! It was life changing for me. I have been stuck on this problem for days and days. I have completed my work now and the weight has been lifted off my shoulders...THANKS AGAIN!!!

-Original poster :)

Obtuse scalene triangle.

Sides: a = 183.363   b = 122.242   c = 62.867

Area: T = 1100.179
Perimeter: p = 368.473
Semi-perimeter: s = 184.236

Angle ∠ A =163.362° = 163°21'45″ = 2.851 rad
Angle ∠ B =11.004° = 11°15″ = 0.192 rad
Angle ∠ C =5.634° = 5°38'1″ = 0.098 rad

Height: ha = 12
Height: hb = 18
Height: hc = 35 - this is the largest positive integer for the value of CF

a=listfor(n, 1, 3, (9+n) nCr 4);print a, "=", sum a

(210, 330, 495) = 1035 - total number of solutions.

Acute isosceles triangle.

Sides: a = 27.713   b = 27.713   c = 27.713

Area: T = 332.554 - area of the triangle ABC.
Perimeter: p = 83.138
Semi-perimeter: s = 41.569

Angle ∠ A =60° = 1.047 rad
Angle ∠ B =60° = 1.047 rad
Angle ∠ C =60° = 1.047 rad

Height: ha = 24
Height: hb = 24
Height: hc = 24

Do not post A O P S homework.  That is cheating.

I'll take you through an alternative method, but leave you to fill in the detail.

Referring to the diagram given in the question, start by drawing a line through C and the point of contact of the cube with the plane, (call that point D), onto its point of intersection with the line AB. Call that point E.

Draw a diagonal across the top face of the cube from OC to the point of contact with the plane. Call the point on OC,  F.

You now have two (vertical) similar triangles, COE and within that the smaller CFD.

Let the side length of the cube be h and let k be the length of OE

You should then be able to use the similar triangles to show that 

\(\displaystyle h = \frac{ck}{(c\sqrt{2}+k)}.\)

Moving on to the base triangle AOB, the angles AOE and BOE are both equal to 45 deg making it easy to write down the areas of the triangles AOE and BOE. (Area AOE = aksin(45)/2.)

Now say area AOB = area AOE + area BOE and you should be able to show that   \(\displaystyle k= \frac{ab\sqrt2}{(a+b)}.\)

Finally, substitute that into the expression for h and tidy up.

OHHHHH I think I just figured it out.  Would it be like A(x+y)?  So then it would be x+3y?

The answer is 12*sqrt(7).

XY = 14.

If the 5 vowels are allowed to repeat, then you should have:

5 x 26^3 ==87,880 - "words"

If the vowels are NOT allowed to repeat, then you should have:

5 x 21^3 ==46,305 - "words"

Note: The above permutations are for 4-letter "words". For 3-letter "words", there are:

5 x 26^2 ==3,380 - "words"  and  5 x 21^2 ==2,205 - "words"

Oh yes, I see....Thanks a mill Alan