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c=0;a=listfor(n, 1, 15, (listforeach(b,(80/27), c=3#(b + c)) 

OUPUT = (1.436289793, 1.638549773, 1.663285786, 1.666260862, 1.666617969, 1.666660823, 1.666665965, 1.666666583, 1.666666657, 1.666666665, 1.666666667, 1.666666667).

Note: As you can see, it converges very rapidly to = 1 2/3

wow thx a lot! I guess i had it in my mind that it had to be a 30-60-90 triangle... THANK YOU SO MUCH!!!

it is a cube so the sides are the same length....

s^2 + s^2 = diagonal ^2

2 s ^2 = (5 sqrt2)^2

s^2 = 25

s = side length = 5         volume would   be    5 x 5 x 5 = 125 units³

1/3 base x height = pyramid volume   

     base is a 3-4-5 right triangle    area = 1/2 * 3 * 4  = 6

 1/3 (6) (6) = 12 ft³         

sum_(n=1)^99(3 n^2 + 3 n + 1) = 999,999

If there are 9 girls ahead of him , therre must be 7 boys ahead of him and then Kamal  all of the rest of the boys are after him

   now find how many boys are in class

b = boys    2b = girls     

b   + 2b = 60

       b = 20  

so there is 7 ahead + kamal = 8  in the top 17                       which leaves  20 - 8 = 12 after Kamal

22   boys to choose 2      22 c 2 =   22! / (20!2!) = 231

      TIMES

14   girls to chose 2          14 c 2 = 91

231 x 91 = ........   

You can write T = 1/(1 - r)^2.

There is missing information here!

3 and 5 make 8, which in base 6 is 2.

As follows:

1*3+2*5+3*7+4*9+5*11+6*13+7*15+8*17+9*19+10*21+11*23+12*2 = 1102

i.e. there are 3 ones, 5 twos, 7 threes, ... 23 elevens and 2 twelves.

There are two solutions, R and r. Evaluate R - r.

Hello Guest!

The distance \(\overline{AB}\)  intersected by the tangent is divided into a and b.

\(a:(6-a)=r_A:r_B\\ a:(6-a)=1:2\\ 2a=6-a\\ a=2\\ b=4\)

\(r_A:(r_A+R)=a:\overline{AB}\\ 1:(1+R)=2:6\\ 6=2+2R \)

\(R=2\)

The radius of the third circle is 2.

There are not two solutions R. 

  !

54 + 65 - 12 = 119 - 12 = 107

Hope this helps Guest,

I suspect "strictly inside" excludes those on the boundary.

Here's an alternative method.

\(\displaystyle abc = 15, \text{ so, }abcd=15d,\text{ so, }\frac{abcd}{15}=d \quad \dots\quad(1).\)

Similarly,

\(\displaystyle \frac{abcd}{30}=a,\quad \dots\quad (2),\\ \displaystyle \frac{abcd}{10}=b,\quad \dots \quad(3),\\ \displaystyle \frac{abcd}{6}=c,\quad \dots \quad (4).\)

Adding  (1), (2), (3) and (4),

\(\displaystyle a+b+c+d=abcd\left(\frac{1}{15}+\frac{1}{30}+\frac{1}{10}+\frac{1}{6}\right)=abcd\left(\frac{11}{30}\right)\quad \dots \quad (5). \)

Multiplying the original four equations,

\(\displaystyle a^{3}b^{3}c^{3}d^{3}=15\times 30\times10\times6=2^{3}\times3^{3}\times5^{3},\\ \text{ so }abcd=2\times 3 \times 5 = 30.\)

Then, from (5),

\(\displaystyle a+b+c+d=30.\left(\frac{11}{30}\right)=11.\)

A lattice point is a point whose coordinates are integers. How many lattice points are strictly inside the triangle formed by the points (0,0), (0,7), and (8,0)?

Hello Guest!

\(f(x)=-\frac{7}{8}x+7\)

lattice points: 8+7+6+5+4+3+2+1+1

There are 37 lattice points in the triangle.

  !

A boat is 130 m from the base of a cliff. The cliff is 80 m high. Determine, to the nearest degree, the angle of depression of the boat from the top of the cliff.

Hello Guest!

\(tan(\delta )=\frac{130m}{80m}=1.625\\ \delta =arctan (1.625)\)

\(\delta \approx 58^0\)

  !

A spherical orange is cut about the vertical axis into 8 equal slices as shown above.

What is the ratio of the total surface area of the 8 slices to that of the original orange?

Hello Guest!

\(S_{slices}=4\pi r^2+4\times \pi r^2\\ S_{orange}=4\pi r^2\\ \color{blue}S_{orange}:S_{slices}=1:2\)

  !

M is the midpoint of PQ and N is the midpoint of PR, and O is the intersection of QN and RM, as shown. If QN is perpendicular to PR, QN = 12, and PR = 14, then find OR.

The ratio of ON to QO is 1/2 or 1 : 2

If QN = 12 then ON = 4

If PR = 14 then NR = 7

We have the right triangle ONR

OR = sqrt ( ON² + NR² ) = 8.062257749   

As follows:

Find all complex numbers z z^2=i

1)  \(i=\sqrt {-1}\)

2) Euler:

 \(e^{i\pi}=-1\\ \sqrt{e^{i\pi}}=\sqrt{-1 } \\ e^{i \frac{\pi}{ 2}}=i\)

\(\sqrt{ e^{ i \frac{\pi}{2} } } =\pm \sqrt{i} \\ e^{ i\frac{\pi}{4} } =\pm \sqrt{i} \\ \sqrt{i} = \pm e^{i\frac{\pi}{4} } \)

\(z^2=i \\ z= \sqrt{i}\)

\(z_1= + e^{i\frac{\pi}{4}} \\= \cos(45^\circ) + i \sin(45^\circ) \\ = \frac{\sqrt{2}} {2}+i \frac{\sqrt2} {2} \)

\(z_2= -e^{i\frac{\pi}{4}} \\=- \frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\)

A   Are they are limited to the amount of paint they have?      80g / 1 g/mile  = 80 miles       using all of the paint

          50 trucks /2 trucks/mile  = 25 miles     using all of the  trucks of asphalt

B    25 miles of bridge =  =    25 trucks   leaving 25 trucks of asphalt         25 trucks/2 trucks/mile = 12.5 miles of land highway  possible

       25 miles of bridge =       25 miles x 3 gal/mile = 75 gallons of paint    leaving only 5 gallons of paint for 5 add'l land miles

          25 miles of bridge + 5 land miles = 30 miles total highway .

Hi, Geno! Please take the time and read the question again. 

A circle of radius 12 is cut out of a piece of paper. A 90-degree sector is then cut from this circle. This 90-degree sector is rolled into a cone. What is the volume of the cone?

r = 3        h = 11.619

V = pi * r² * h / 3 ≈ 109.51 u³  

Draw a picture to help visualize this:

Find out how far apart they are     tan 27 = 40/d      d = 78.5 m

Then the height above the 40 m building   is  tan21 = height /78.5    

             height above 40 m   = 30.133 m

                      and add the 40 m height to this = ~~ 70.1 m

d-8 = 2/3 (d+13)     (given)

1/3d =  26/3 + 8

d = 26 + 24

d = 50 y/o now

There are 252 numbers between 100 and 999 that have AT LEAST one 9 in them. If you include all repeated digits such as 199, 999, then there are 280 numbers between 100 and 999 that have 9 in them.