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Why do you need the answer so quickly? Is it for a test or quiz?

-Grace

Let F  =(0,1)     C  =  (2,2)

And  2CG  = 3 FG     means  that

FG / CG = 2/3

So....there are 5 parts of  FC  and FG  is 2 of them....thus  FG = (2/5)FC

We can find the coordinates of  G    thusly

[ 0 + (2/5) (2 -0)  ,  1 + (2/5)(2 - 1) ]   =     (4/5, 1 + 2/5 ) = (4/5, 7/5)

And 7/5 is the height of EGB  and the base EB  = 1

So  [EGB ]  =(1/2) (1) (7/5) =   7 /10

Let B  =(0,0)   C = (8,0)

And  we can find the x coordinate of  A by letting the  circles   x^2 + y^2  = 25   and  (x - 8)^2 + y^2 = 49  intersect

Subbing    y^2  = 25 -x^2  into  the second equation we have that

(x - 8)^2  + 25 - x^2  =  49

x^2  - 16x + 64  + 25 - x^2  = 49

-16x + 89  = 49

16x  = 40

x = 40/16  = 2.5  ⇒    x^2  = 6.25

And  y = sqrt (25 - 6.25)  = sqrt [ 18.75 ]   = sqrt (75/4)  = sqrt (25 * 3) / 2  = (5/2) sqrt (3)=  the altitude of this triangle

The area of this triangle  =   (1/2)( BC) (altitude) =  (1/2) (8) (5/2) sqrt (3)  = 10sqrt (3)

We can find the radius,R, of the  inscribed circle  thusly

(1/2) R  ( 8 + 7 + 5 )  =  10sqrt (3)

10R  =  10sqrt (3)

R =  sqrt (3)

We  are counting sets and orders within those sets....a permutation  will do this.....

The number of ways  is P(20,3)   = 6840 

What is the radius of the circle inscribed in triangle ABC if AB = 5, AC=7, BC=8? Express your answer in the simplest radical form.

r = √3        area of ΔABC = 10r         area of a circle is  3pi   

great job both of you!

25 total letters (ni 'y's)     a e i o u     20 consonants and 5 vowels

   20  x  5  x  20 x  5     (This assumes letters can appear more than once)

      = ..........

The question is how to show that the above two products are equal to each other.

I don't know if this is the best way, but this is how I would do it:

\(\large {\prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1+m}^{k=n+m}a_k\)     because  \( \prod\limits_{k=1+m}^{n+m}a_k\)   means the same as  \( \prod\limits_{k=1+m}^{k=n+m}a_k\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_k\)      by subtracting  m  from both sides of each equation

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_{(k-m+m)}\)         since  k - m + m  =  k

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{{\color{RubineRed}k-m}=1}^{{\color{RubineRed}k-m}=n}a_{({\color{RubineRed}k-m}+m)}\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{k=n}a_{(k+m)}\)       because we can replace the pink text with whatever we want

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{n}a_{(k+m)}\)

_

Thanks Grace  :)

1  -  The mmi =  1 [mmi - stands for "Modular Multiplicative Inverse" ]
5  -  The mmi =  5
7  -  The mmi =  7
11  -  The mmi =  11
13  -  The mmi =  13
17  -  The mmi =  17
19  -  The mmi =  19
23  -  The mmi =  23

Count them and you get a == 8

Please do not edit your question if it has been answered.

And even if it has not been answered yet, please explain what you are changing.

I remember this

\(log_{10}100=2\qquad because \qquad 10^2=100\)

A log is a power

so

\(if \;\;x=log_3 18\;\;then \;\;18=3^x \)

I will use change of base log law to solve this

\(log_y x = \frac{log_a x}{log_a y}\)

\(log_318=\frac{log_{10}18}{log_{10}3}\approx 2.6309\)

So it will only 'equal' 3 if it is rounded to the nearest whole number.

Let  the eqaution of the  circle  be   x^2 + y^2   = 2.5^2   ⇒  x^2 + y^2   = 6.25

Call  the side of the  purple square, a

Then,  when y = a ,  we can find  x  as    sqrt ( 6.25  - a^2 )

And using symmetry,   the distance from  P  to this point  = 2 sqrt (6.25 - a^2)

Using the secant-tangent theorem, we  have  that

a^2  = [ (1/2)a ] [( (1/2)a  + 2sqrt ( 6.25 - a^2 ]

a^2=  (1/4)a^2  +  a sqrt (6.25 -a^2)

(3/4)a^2  =  a sqrt ( 6.25 -a^2)    divide out a

(3/4)a  = sqrt (6.25 - a^2)        square both sides

(9/16)a^2   = 6.25  - a^2

(9/16)a^2  =  25/4 - a^2

(25/16)a^2  =  100/16

25a^2   =100

a^2 = 100/25  =  4

a  = 2

So....the area of the pink square  = a^2   =  2^2   = 4

And the area of  the orange square is  [ (1/2) a]^2  = [(1/2)(2)]^2  =  1

And the area of  the   aqua suare is  ( 5 -2)^2   =3^2   = 9

So....the total area is    4 + 1 + 9  =   14

Let  the center of the large circle  be O  and let the  centers of the  the smaller circles be M and  N

Connect  OM, ON  and MN

OM  = ON  =12 - 4   = 8

MN =  $ + 4   = 8

So....we  have an equilatera triangle  OMN.....this means that angle MON  = 60°

And   OA   =12  = OB

So....by the Law of  Cosines we can find AB  thusly

AB^2   =  OA^2  + OB^2  - 2 (OA * OB)  cos (60°)

AB^2  = 12^2  + 12^2  -  2(12*12) (1/2)

AB^2   = 288  - 144

AB^2  =  144   take the positive square root

AB  =12

We might also see  that  since OA and OB  = 12  and angle AOB  also = 60°, then triangles OAB and OMN  are similar...so  triangle OAB is also equilateral....so AB  =   12

x ( x + 5) =   -n

x^2  + 5x  +  n    =  0

For us to  have a real solution here, the discriminant  must  be  equal or  greater to  0

So

5^2 - 4(1) n   ≥  0

25  - 4n  ≥  0

Note that  this is true when  n  is an  integer from  1 to 6  inclusive

So....we will have  no  real solutions when n=  7,8,9 or 10

So

P(no real solutions)  =  4/10   = 2/5

Treating the problem as a calculus max/min problem gets the result easily enough, but in keeping with the suggestion in the question here's the Cauchy-Schwarz solution.

For n = 3, Cauchy-Schwarz says that

\(\displaystyle a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \leq(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{1/2}(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})^{1/2},\)

so,

\(\displaystyle 1 =a+b+c=a.1+(\sqrt{2}b)(1/\sqrt{2})+c.1\leq(a^{2}+2b^{2}+c^{2})^{1/2}(1+\frac{1}{2}+1)^{1/2},\\ 1\leq(a^{2}+2b^{2}+c^{2})^{1/2}(5/2)^{1/2},\)

so squaring and cross-multiplying,

\(\displaystyle a^{2}+2b^{2}+c^{2}\geq2/5.\)

Let's see if  we  have some sort of repeating pattern here

First composition

f(12)  = 13/11

g (f(12)) = g ( 13/11) = 11/13

Second composition

f(11/13)  =  12

g (12)  = 1/12

Third composition

f(1/12)  =  13/11

g(13/11)  =   11/13

The pattern repeats afer 2 compositions  .......so ater 16 compositions we will end up with  1/12

2.

Possible base 5  integers   are   1(5)^2  + 0 + 0   = 25 base 10  to 4(5)^2 + 4(5)  + 4   = 124  base 10

Possible  base 8 integers  are   1(8)  + 0  =  8  base 10  to 7(8) + 7 =  63 base 10 

So....the intersection  of  these (in base 10 )  is  [  25, 63]

The average  of  these integers   = 

[ 25 + 63 ]  / 2  =  

44

Smallest two digit base 8 number  1 0      =  8₁₀

Largest two digit base 8 number    7 7     = 63 ₁₀

 Smallest 3 digit base 5    100 = 25₁₀

Largest three digit base 5 = 444 =124₁₀

So we need the average of base 10 numbers from 25   thru 63       1716/39 = 44 average₁₀

Set  2x + 8  = -5

       2x  =  -13

         x = -13/2     this is in the domain, so it is  one solution

Set  -x^2  - 3x  + 1   =  -5

         -x^2  - 3x + 6  =     0    multiply through by -1

          x^2 + 3x - 6  =  0         complete the square on x

          x^2  + 3x + 9/4  =  6 + 9/4

          ( x + 3/2)^2   = 33/4      take  both roots

            x + 3/2 = ± sqrt (33)/2   

So

x=  sqrt (33)/2  -3/2  ⇒   ≈ 1.37     this  is in the  domain

x  = -sqrt (33)/2  - 3/2  ⇒ ≈ -4.37   this is not in the  domain    

So....the sum of the  values   =   

 -13/2  - 3/2  +  sqrt (33)/2  =  

-16/2  + sqrt (33) /2  =

sqrt (33)  - 16

___________

         2

OP here.

I actually managed to get this problem! Here, a = c = 2/5 and b = 1/5.

the expression goes to 2/5. this was scary! thank you!

Let O  be the center of the circle

We can split this irregular pentagon into  5 triangles

Triangles  AOB , BOC  and  DOE  are all equilateral with sides  of 2

Their  combined areas   are  

3 (1/2)  ( 2)^2  sin 60°  =  6sqrt (3) /2  = 3sqrt (3)

And triangles   COD  and EOA   each have  sides of  2  and  an included angle of  [ 360 - 3(60)] / 2   =

90°

Their combined areas   =  2* (1/2) (2)^2 sin (90°)  =  4

So

[ABCDE ]   =   4 + 3sqrt (3)  (units^2)

These were my first steps that I tried applying to the problem, too. However, even though it seems intuitively true, it is not the correct answer. Perhaps values other than 1/3 would result in lower numbers? 

That's where I'm stuck. I can't find anything else that could be lower. 

Thank you for your time!

10! = [2^8 * 3^4 * 5^2 * 7] mod 2^7 ==0 - The remainder.

10!  = 1 * 2 * 3  * (2 * 2)  *  5  * ( 3 * 2)  * 7 * ( 2^3)  * 9  *  (5 * 2)  / 2^7   =

2^8   (  3 * 5 * 3 * 7 * 9 * 5)                       

________________________   =          2 ( 3^2 * 5^2 * 7 * 9)   =    28350 

             2^7

(Remainder  =  0 )