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First off, let's let \(DC = 3BD\)

We can write two equations off the problem

\(AD = \sqrt{AC^2 - (3BD)^2 } = \sqrt{15^2 - 9BD^2 } \\ AD = \sqrt{AB^2 - BD^2} = \sqrt{10^2 - BD^2 }\)

Notice that we have AD on both equations, so we can equate the two to get

\( \sqrt{15^2 - 9BD^2} = \sqrt {10^2 -BD^2 }\)

Now, we just have to solve for BD^2. 

\(15^2 - 9BD^2 = 10^2 - BD^2 \\ 225 - 9BD^2 = 100 - BD^2 \\ 125 = 8BD^2 \\ 125/8 = BD^2\)

We know that 

\(AD =\sqrt{10^2 - BD^2}\) so we can just plug in BD^2 to find AD. 

\(AD = \sqrt{100 - 125/8} = \sqrt { 675/8} = \frac{15\sqrt3 }{2 \sqrt2}\)

This rounds to about 9.19. 

Thanks! :)

From the problem, we can write the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15 \)

 where t is toast, b is bagel and m is muffin. 

From the first two equations, we have

\( [3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Subbing this in to the third equation, we have 

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is about 262 cents. 

Thanks! :)

Not too bad of a problem. 

\( \frac{b^2 - a^2}{ b - a } = \frac{ (b -a) (b + a)}{(b -a)} = b + a = 2\)

Thanks! :)

We knowt hat AK is the altitude, so we have

\(AK = \sqrt{AC^2 - CK^2} = \sqrt{8^2 - 3^2} = \sqrt{55}\) from the pythagorean theorem. 

We can now find AB, with

\(AB = \sqrt{AK^2 + BK^2} = \sqrt{ 55 + 4 } = \sqrt { 59 } \)

so AB is \(\sqrt{59}\)

Thanks! :)

Well, unfortunately, this triangle can't ever exist. 

Using the Law of Cosines, we have

\(AC^2 = CM^2 + AM^2 - 2 (AM * MC)cos (AMC) \\ AB^2 = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC) \\ 17^2 = x^2 + 12^2 - 2 (AM * x) cos (AMC) \\ 7^2 = x^2 + 12^2 + 2(AM * x) cos (AMC)\)

Adding these two equations together, we get 

\(338 = 2x^2 + 288 \\ 50 = 2x^2 \\ 25 = x^2 \\ 5 = x \\\)

Alright. 

Now, we have \( BM + CM = 5 + 5 = 10 = BC\). 

However, according to the Triangle Inequality Theorem, we have

\(AB+BC>AC \\ AB +BC > 17\)

However, we have

\( AB + BC = 7 + 10 = 17\)

Meaning this is an invalid triangle. 

Thanks! :)

At a cafeteria,   

Mary orders two pieces of toast and a bagel, which comes out to $3.15.   

Gary orders a bagel and a muffin, which comes out to $3.50.   

Larry orders a piece of toast, two bagels, and three muffins, which comes out to $8.15.   

How many cents does one bagel cost?   

                                           Mary          2T + B = 315 ............. —>  T = (315 – B) / 2       

                                           Gary          B + M = 350 ................—>  M = (350 – B)      

                                           Larry          T + 2B + 3M = 815  

                                           Substitute for Toast & Muffin into Larry's order  

                                           [(315 – B) / 2]  +  2B  +  [3 • (350 – B)]  =  815   

                                           [(315 – B) / 2]  +  2B  +  (1050 – 3B)  =  815   

Multiply both sides by 2       (315 – B)  +  4B  +  (2100 – 6B)  =  1630    

Combine like terms              –3B + 2415  =  1630     

                                                        –3B  =  – 785     

                                                            B  =  261 ²/₃  ¢    

They usually construct these problems to come out with an even number.   

I've gone back over the steps a hundred times and cannot find a mistake.    

_.

We don't actually have to find the area of the polyhedron itself. 

We know how to find the area of the cube and the area of the pramids cut off, so we just have to subtract them. 

Each of the pyramids cut off must have a right iscocoles base with sidelengths 3 and height 3. 

From this, we can write

\(\text{Volume} = 6^3 - 8 \cdot \dfrac13 \left(\dfrac{3 \cdot 3}2\right)\cdot 3 = 180\)

So the volume is 180. 

Thanks! :)

Well, I don't reccomend it, but I used a calculator because I'm too lazy. 

\(729y^{12}+2916y^{11}-2430y^{10}-19980y^9+135y^8+59976y^7+6364y^6-99960y^5+375y^4+92500y^3-18750y^2-37500y+15625\)

As you can see, the coefficient for y^4 is 375. 

Thanks! :)

There's a pretty handy formula to find the circumradius of a triangle. 

We have \(\frac{(abc)}{\sqrt{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}}\) where a, b, and c are the sidelegnths. 

Here, we just have \(\frac{a^3}{\sqrt{3a^4}} = \frac{a^3}{a^2 \cdot \sqrt3}= \frac{a}{\sqrt{3}}\). 

So \(\frac{a}{\sqrt3}\) is our answer. 

Thank You! :)

This was the post that got me to 300 points LOL :)

Inradius =    Area / semi-perimeter  =   (1/2)a^2 * sqrt (3) / 2            (sqrt (3) / 4) a^2

                                                             __________________ =     ______________ =

                                                                        (3a)/2                             (3a/2)

sqrt (3) a

_______

      6

             A

       14-x     14

     Z                 Y

    x                       8

B                              C

We already found that BC =  8

Since CZ is a bisector

BZ / BC  = AZ / AC

x / 8  = (14 -x) / 22

22x = 8 (14 -x)

22x  =  112 - 8x

30 x  = 112

x = 112/30  =  56 / 15  =  BZ 

                 A

                    14

    14                        

         Z               Y

                            8

B                              C

Because BY  is a bisector

BC / CY  =  AB / AY

BC / 8  =  14 / 14

BC =  8 * 14 / 14  =  8

         B

    6         8

A                   C 30

Law pf Cosines

AB^2  = AC^2 + BC^2  - 2(AC * BC) cos (30)

6^2 = AC^2 + 8^2  -  2 (AC * 8C) sqrt (3)/2

36  - 64 = AC^2   - 8sqrt (3) AC

-28 = AC^2 - 8sqrt (3) AC

AC^2  - 8sqrt 3 AC  + 28 =  0       solving for AC

AC   ≈ 2.46    or  AC ≈  11.4

So..because pf the SSA situation.....two possible areas result

[ ABC] =   (1/2) (2.46) (8) (1/2)  ≈ 4.92

[ABC ]  =  (1/2) (11.4) (8)(1/2)  ≈  22.8

The system is inconsistent

From the second equation , r =1

From the last equation r  = 9

Add the equations

2a = 10

a = 5

5 ( b + 1) = 250

b + 1 =  250 / 5

b + 1  =  50

b = 49

\(\frac{1}{64a^3 + 7} - 7 = 0 \)

1/[64a^3 + 7]   =  7

1 =  7 [ 64a^3 + 7]

1 = 448a^3 + 49

-48  = 448 a^3

[ -48 /448 ] = a^3

-3/28 = a^3

a =   - ∛ ( 3/28)

British Flag Theorem

AX^2 + CX^2  =  BX^2 + DX^2

3^2   + 3^2  = 3^2 + DX^2

3^2 = DX^2

3 = DX

ab^4 = 48           divide  the first wquation by the second

ab^2 = 4

b^2 = 12          take the both roots

b =  ± sqrt (12) = ± 2sqrt (3)

ab^2 =4

a (12) = 4

a = 4/12 = 1/3

Let x be the length from the right edge to the strip. 

We have

\([rectangle]=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ [strip]=1\cdot \sqrt{8.5375^2+8^2}\\ [strip]=11.7\)

Thanks! :)