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b)   Take 1/5 th of the pentagon as shown below

     we need to find 'b'   (which is 1/10 th of the perimeter)

         calculate the area of 10 of these triangles = 6 

              10 *  1/2 b h

              10 * 1/2 ( hyp sin36 * hyp sin54)  = 6

                             hyp^2 sin36 sin 54 = 1.2

                             hyp = 1.588556        then   'b' = hyp * sin 36 = .9337296      perimeter = 10 * b =  9.337296

Now do the same for the larger pentagon   to   get tthe ratio of the perimeters   (basically do the above calculations wth '6' changed to 73.5)    

Starts with 'x'          Sells  1/9    then has  8/9 left 

  sells 3/4 of this leaving 1/4 of this     so now he has   1/4 ( 8/9) x

     adds 200     1/4 ( 8/9) x + 200   = x + 32

                                   x= 216

(7.8, 2.2)

Please write your question in English

It can also be found here: https://web2.0calc.com/questions/math-problem_35879

There are 400 numbers that are divisible by 5 within the range. Of these, 133 of them are divisible by 15, so your answer is 400-133 = 267.

Point D is on side BC of triangle ABC. We know ABC ~ ACD and angle A = 49 degrees. If BC = AD, what is angle B in degrees?

Find a monic cubic polynomial P(x) with integer coefficients such that $P(\sqrt[3]{4} + 1) = 0$ . (A polynomial is monic if its leading coefficient is 1.)

\(P(x)=[x-(\sqrt[3]{4}+1)][x^2+ax+b]\\ P(x)=[(x-1)-\sqrt[3]{4}][x^2+ax+b]\\ \)

Remembering that    \((p^3-q^3)=(p-q)(p^2+pq+q^2)\)

\(Let \;\;p=x-1    \quad and  \quad  q= \sqrt[3]{4}\\ then\\ (x-1)^3- (\sqrt[3]{4})^3=[(x-1)-\sqrt[3]{4}][(x-1)^2+\sqrt[3]{4}(x-1)+4^{2/3}]\\ and\\ (x-1)^3- (\sqrt[3]{4})^3=(x^3-3x^2+3x-1)-4\\ (x-1)^3- (\sqrt[3]{4})^3=x^3-3x^2+3x-5\\ so\\ P(x)=x^3-3x^2+3x-5\\\)

A written response will be appreciated.

LaTex:

Let \;\;p=x-1    \quad and  \quad  q= \sqrt[3]{4}\\
then\\
(x-1)^3- (\sqrt[3]{4})^3=[(x-1)-\sqrt[3]{4}][(x-1)^2+\sqrt[3]{4}(x-1)+4^{2/3}]\\
and\\
(x-1)^3- (\sqrt[3]{4})^3=(x^3-3x^2+3x-1)-4\\
(x-1)^3- (\sqrt[3]{4})^3=x^3-3x^2+3x-5\\
so\\
P(x)=x^3-3x^2+3x-5\\

A=7,   B=11,   C=5   and   D=13

7  + 2 =11 - 2 =5 + 4 =13 - 4 ==9

Their product =7  x  11  x  5  x  13 ==5,005

You did all the work catmg :)

Normally I would not have answered this on athis new thread, but the original unintentioned question was a completely different question.

\(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\\ =\frac{4}{27}\left( \left[ 2^{\frac{3+m}{3}} \right] + \left[ 2^{\frac{m}{3}} \right] \right)^3\\ =\frac{4}{27}\left(\left[ 2^{\frac{3+m}{3}} \right]^3 + 3*\left[ 2^{\frac{3+m}{3}} \right]^2 \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] ^2 + \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+2m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{2m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+3m}{3}} \right] + 3*\left[ 2^{\frac{3+3m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{2+m} \right] + 3*\left[ 2^{1+m} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4*2^m}{27}\left(\left[ 2^{3} \right] + 3*\left[ 2^{2} \right] + 3*\left[ 2 \right] + \left[ 1 \right] \right)\\ =\frac{4*2^m}{27}\left(8+ 12 + 6 + 1\right)\\ =\frac{4*2^m}{27}\left(27\right)\\ =2^{(m+2)} \)

LaTex

\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\\

=\frac{4}{27}\left(     \left[  2^{\frac{3+m}{3}} \right]    +    \left[ 2^{\frac{m}{3}} \right]    \right)^3\\
=\frac{4}{27}\left(\left[  2^{\frac{3+m}{3}} \right]^3 
+ 3*\left[  2^{\frac{3+m}{3}} \right]^2   \left[ 2^{\frac{m}{3}} \right]     
+ 3*\left[  2^{\frac{3+m}{3}} \right]   \left[ 2^{\frac{m}{3}} \right]  ^2  +  \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{\frac{6+2m}{3}} \right]   \left[ 2^{\frac{m}{3}} \right]     
+ 3*\left[  2^{\frac{3+m}{3}} \right]   \left[ 2^{\frac{2m}{3}} \right]   +  \left[ 2^{m} \right]  \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{\frac{6+3m}{3}} \right]      
+ 3*\left[  2^{\frac{3+3m}{3}} \right]     +  \left[ 2^{m} \right]  \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{2+m} \right]      
+ 3*\left[  2^{1+m} \right]     +  \left[ 2^{m} \right]  \right)\\

=\frac{4*2^m}{27}\left(\left[  2^{3} \right] 
+ 3*\left[  2^{2} \right]      
+ 3*\left[  2 \right]     +  \left[ 1 \right]  \right)\\

=\frac{4*2^m}{27}\left(8+  12 + 6    + 1\right)\\
=\frac{4*2^m}{27}\left(27\right)\\
=2^{(m+2)}

What has -pi/2 got to do with anything.

Please put the question altogether properly.

360 / 100 = 3.6

3.6 * 250 = 900

circle area = 900pi

circle radius = √900

Attn: Ora2004

You ask questions but you have never once responded to an answer.

I would not encourage people to answer you.  You are totally ungratful and are quite likely cheating.

If you were trying to learn you would be interacting with the answerers.

The question is what is 10c and 10d.

What is the question?

ok  :)

Where was that?

If you want me to see it you need to incluse a link.

(I should hae left a link too.)

Solved here: https://web2.0calc.com/questions/geometry_69190

Already answered here: https://web2.0calc.com/questions/1-the-graph-of-x-3-2-y-5-2-16-is-reflected-over-the-line-y-2-the-new-graph-is-the-graph-of-the-equation-x-2-bx-y-2-dy-f-0-for-some-constants

Yes, I saw that. Btw, on my last problem, in the second sentence I tried to write -pi/2

Well, tbh I just plugged it into the calculator on this website lol, then divided the radians amount by pi. 

Did you see my edit on our last post.  The answer has to be positive.  Take  a look.

Well .. are you going to show us how you did it?   OR guide us ??