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answer is 5f%?

I don't know

da answer is 6^7a = 5f%

Ah! I realized that the first terms starts as 1. So using the same strategy.

If we count up 5 in the pattern, we will end up with the last digit of term 36.

So we count back 1 in the pattern

the correct answer is 8!

Let the number of girls  =  g

Let the number of boys  =  b

b + g  =  20

b  =  20 - g

probability that a girl is chosen  =  \(\frac23\)( probability that a boy is chosen )

\(\frac{g}{20}\,=\,\frac23(\frac{b}{20})\\~\\ \frac{g}{20}\,=\,\frac23(\frac{20-g}{20})\\~\\ g\,=\,\frac23(20-g)\\~\\ \frac32g\,=\,20-g\\~\\ \frac32g+g\,=\,20\\~\\ \frac52g\,=\,20\\~\\ g\,=\,8\)

_ _ _ _ _ _

it's not 2..

A similar question was posted a long time ago

the first few terms are 1, 2, 2, 4, 8, 32, 256, 8192, 2097152...

Notice how the last digits are forming a pattern, (2, 2, 4, 8, 2, 6, 2, 2, 4, 8, 2, 6, 2...

The pattern is 2 -> 2 ->  4 -> 8 ->  2 -> 6

So if the pattern repeats every 6 terms. And assuming you are trying to find the 35th term. You divide 35 by 6 and find the remainder, which is 5

So count up 5 terms in the pattern.

[fixed]

we count up 4 terms in the pattern because the first term is 1

so 8 is the answer

Someone check this.

Note: In these questions, always check for a pattern!!!

We need to figure out the denominator first of the probability fraction

100-999

That is 900 digits

Ok so our denominator is 900

Now we need to figure out the numerator. We do this by finding all the three digit numbers divisible by 11

We try listing (I couldn't find any better way to do this

110

121

132

143

154

165

176

187

198

----

209

220

231

242

253

264

275

286

297

----

Notice how the first number's last two digits start from 10 then decreases to 9 at 200

notice how the last number last two digits start from 98 and decrease to 97 at 200

This means each hundred we have 9 numbers divisible by 11

This means that there are 9*9 (9 different hundreds) = 81 numbers 

We have the numerator which is 81 and the denominator which is 900

Simplifying we have \(\boxed{\frac{9}{100}}\)

how could both the last two answers be correct?

\(a^3 + b^3 = (a+b)(a^2 -ab +b^2)=\\ (a+b)^2 = a^2 + b^2 + 2ab=15+2ab=25\\ 2ab=10\\ ab=5\\ a^3+b^3=5(15-5)=50\)

\(\text{Suppose $p(x)$ is degree 1 or less and has $1-i$ as a root}\\ c_0+c_1(1-i) = 0\\ c_0+c_1 = i c_1\\ \text{The left side is the sum of two real numbers and thus real.$\\$ The right side is the product of a real number and $i$ and is thus $\\$ imaginary so they cannot be equal..$\\$ Thus $p(x)$ must be of degree 2 or greater}\)

The degree of the polynomial determins how many roots it has.

If you have the linear polynomial \(ax+b\) and the cubic polynomial \(cx^3+dx^2+ex+f\), the highest power in the product \((cx^3+dx^2+ex+f)(ax+b)=acx^4+(ad+bc)x^3+(ae+bd)x^2+(af+be)x+bf\) is \(x^4\) so the polynomial will have 4 roots.

The height of an eqilateral triangle  with side, S   = S*sqrt (3) / 2

So  AD  is twice this  =   S*sqrt(3)

So

AD / BC  =     S*sqrt(3)  / S    =   sqrt(3)

OK....let me see if I can spot my error....

3. Expand (a+b)^3

We only need  the 3rd row of Pascal's Triangle for the coefficients  [ note that the first row  = Row 0 ]

So....we have

1a^3  + 3a^2b  + 3ab^2  + 1b^3  =

a^3  + 3a^2b + 3ab^2 + b^3

5.Find the number of paths from  to  that pass through point  (Each step can only go up or to the right.)

Note that a set of moves that get us from  A to C  is  just  (R, R, U, U, U)

And we can choose any 2 of 5 places for R in the set and still get from A to C

So.....the total number of paths from A to C  is  C(5,2)   =  10

Likewise

A set of moves that get us from  C to B  is just (U, R,R,R )

And we can choose any 1 of the 4 places for U in the set and still get from C to B

So....the the total  paths from C to B is just  C(4,1)  = 4

And the total possible paths   =  C(5, 2) * C(4, 1)    =   10 * 4   =    40

thanks so much, Cphil

The second derivative test tells us whether a point is a local maximum/minimum it doesn't tell us anything about global maximum/minimum

like I said above you have to apply the 3 dimensional second derivative test.

The test may prove inconclusive but under certain conditions it guarantees a minimum or maximum as the case may be.

I am seeing a pattern that the critical points obey.

We have the following perimeter

2x + 2y  +  pi x / 2  =   12         divide through by 2

x +  y  +   (1/4)pi x  =  6

(1 + (1/4)pi ) x  + y  =  6

y  =  6  - (1 + .25pi)x

So....the total Area , A, to be maximized is

A  =  x * y  + pi  [(1/2)x]^2/2          substituting for y, we have

A =  x * [ 6 - (1 + .25pi)x] + .25pix^2 /2      simplify

A   = 6x - (1 + .25pi)x^2 + .125pix^2

A = 6x - x^2 - .25pix^2 + .125pix^2

A = 6x - x^2 - .125pix^2

A= 6x -  ( 1 + .125pi)x^2

This might be most easily solved with a graph: https://www.desmos.com/calculator/brhi8ngjzf

The value of x that maximizes the area ≈  2.154 ft

And y = [6 - (1 + .25pi)(2.154) ] ≈ 2.154 ft

So  the rectangular part  =  2.154 ft x 2.154 ft

Corrected Answer

But the domain of the function is an open set (all triplets of strictly positive numbers) how do you prove that there's a minima to the function in this domain?

Critical points are found by setting the gradient equal to zero.

You then have to determine whether they are minima, maxima, or saddle points using the second derivative test.

That's not exactly trival for 3 variables but there is a method for doing it.