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Here is the question

I have looked at this question but this sentence does not make any sense.  Not to me anyway.

" Points C and D are parallel to AC meets CB extended at E "

How can points be parallel to anything ?????

PQ = 2(RS/2 + XY)

tan (B) = 3 / √8

√x + √y + √z = 10

x + y + z = 42

√(xy) + √(xz) + √(yz) = 29

Hello Guest!

1 + 4 + 5 = 10

1 + 16 + 25 = 42

√(1*16) + √(1*25) + √(16*25) = 29

x = 1    y = 16    z = 25

  !

What is the largest integer \(n\) such that \(3^n\) is a factor of

\(1 * 3 * 5 * \ldots * 97 * 99 * 101 * 103 *\ldots * 197 * 199\)?

\(\begin{array}{|rcll|} \hline && 1*3*5*7*9*11 \ldots *195*197*199 \\\\ &=& \dfrac{ 1*2*3*4*5* \ldots *196*197*198*199} {2*4*6*8*10* \ldots *194*196*198 } \\\\ &=& \frac{ 1*2*3*4*5* \ldots *196*197*198*199} {(2*1)*(2*2)*(2*3)*(2*4)*(2*5)* \ldots *(2*97)*(2*98)*(2*99) } \\\\ &=& \dfrac{ 1*2*3*4*5* \ldots *196*197*198*199} {2^{99}*1*2*3*4*5* \ldots *97*98*99 } \\\\ &=& \dfrac{ 199!} {2^{99}*99! } \\\\ && \boxed{ 199! = \ldots * \left.3^{ \lfloor\frac{199}{3}\rfloor + \lfloor\frac{199}{9}\rfloor + \lfloor\frac{199}{27}\rfloor + \lfloor\frac{199}{81}\rfloor } \right. * \ldots \\ 199! = \ldots * \left.3^{66 +22 +7 +2 }\right. * \ldots \\ 199! = \ldots * \left.3^{97}\right. * \ldots \\\\ 99! = \ldots * \left.3^{ \lfloor\frac{99}{3}\rfloor + \lfloor\frac{99}{9}\rfloor + \lfloor\frac{99}{27}\rfloor + \lfloor\frac{99}{81}\rfloor } \right. * \ldots \\ 99! = \ldots * \left.3^{33 +11 +3+1 }\right. * \ldots \\ 99! = \ldots * \left.3^{48}\right. * \ldots \\ } \\\\ &=& \dfrac{ \ldots * \left.3^{97}\right. * \ldots} {2^{99}*\ldots * \left.3^{48}\right. * \ldots } \\\\ &=& \dfrac{ \ldots * \left.3^{97-48}\right. * \ldots} {2^{99}*\ldots} \\\\ &=& \dfrac{ \ldots * \left.3^{\color{red}49}\right. * \ldots} {2^{99}*\ldots} \\ \hline \end{array}\)

\(\mathbf{n=49}\)

**deleted**

-2 (y² +4y)    + 5

x = -2 (y+2)²  +13         

     this is a sideways parabola  of the form     x = a(y-k)² +h      The vertex is  13,-2

so the line x = k = 13 will touch in only one point

11 111 111 111  is the largest

40^13 mod 85 = 10      

https://web2.0calc.com/questions/attempted-twice-i-can-t-find-the-answer

f(f(x)) = f(b)

f(b) = 4    for the first condition:    x-3 = 4    x=7   this is not <=0   so it does not apply

                for the second condition    5x+2 =4    x = 2/5    which IS greater than 0

                       so b = 2/5

Now do this again

f( x)  = 2/5     for the first condition   x-3 = 2/5    x = 3 2/5     this does not apply because x > 0

                        so the second condition applies 

                               5x+2 = 2/5

                                 x =  -8/5 / 5 = -8/25     this does not apply either

There is no value for  which this is true

Here is a graph of the piecewise function.....you will see no value where it = 2/5   ( the green line does not intersect the graph)

https://www.desmos.com/calculator/qt7spyez1o

Thank you so much for answering my question but could you please elaborate on how you got \((n-k+1)!=(n-k)!(n-k+1)\)

The shaded area looks like a circle. 

Complete the square normally as if the "<=" was an "=" to get the equation in the form (x-h)^2 + (y-k)^2 = r^2. 

Then find the area of the circle using r^2. 

=^._.^=

*** edit ***   answer does not change

The only ab from the set that would be a prime number would be   1 * 5     or   1*7   out of 25 possible products.

x is in the domain when (x + 3)^2 - (x - 8)^2 is nonegative.

(x + 3)^2 - (x - 8)^2 >= 0

Simplify the inequality. :))

=^._.^=

thank you so much!! i had to read through a few times to understand... but this was really helpful!! :)))

Area = height * base = 4 x 7 = 28 = a

Perimeter = 7 + 7 + 5 + 5 = 24 =p      because the side triangles form a 3 - 4 - 5 right triangle

   Add them together to get 52 units

In triangle ABC the angle bisectors are AD BE  and CF, which intersect at the incenter I. If Angle ACB = 38 degrees then find the measure of AIE in degrees.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let angle BAC be 90 degrees

∠IAE = 45º

∠AEI = 90 - 1/2(∠ABC) = 64º

∠AIE = 180 - (∠IAE + ∠AEI) = 71º

Shortcut:::

∠AIE = 90 - 1/2(∠ACB) 

The only ab from the set that would be a prime number would be   1 * 5     or   1*5   out of 25 possible products.

For a line x=k (a vertical line) to intersect a  horizontal parabola at only one point, that point must be the vertex....do you know how to find the vertex?

thank you so much! I really appreciate it! ^^

Why do you say this?

Angle AIB = 111 degrees.