All Answers 357330 Answers

you say that every day, if you are gonna do that post it as OFF TOPIC, posting as math questions is for QUESTIONS that have to do with MATH, not a guy who says "be happy!" every single gosh dang day. please stop.

Let's reverse the process

log  ( a⁴ / [ 3² x ]  )     =

log a⁴  -  [ log (3² * x) ] =

log a ⁴ - [ log 3² + log x ]   =

4 log a  - [ 2 log 3  + log x ]

4 log a  - 2 log 3  - log x

Hi Melody,

thanx for the reply...can you just explain how the "5" became a "4" please?

Hi Melody,

well, it's this:

$4loga-2log3-logx$

 goes to:

$log{a^4 \over 3^2}$

the "-logx"...how do i "see" mathematically that it is supposed to join the $3^2$ ?

I'd go for probability. :)

could you add the number of tablespoons used and in the bottle?

thats not a catch, thats a catch and a half!

JOKES ARE MY THING >:(

but funny anyways 8)

So, I going to denote head to h, body to b, and tail to t.

The tail is equal to the head (9 inches) plus half the body.

$t=9+\frac{1}{2}b$

The body is the size of the head plus the tail.

$b=9+t$

Now, plug the first equation into the second for t.

$b=9+(9+\frac{1}{2}b)$

Simplify.

$b=18+{1\over2}b$

Multiply everything by 2 to get rid of fractions.

$2b=36+b$

Subtract b from both sides.

$b=36$

So, the body is 36 inches long

Now, plug 36 into the first equation.

$t=9+{36\over2}$

Simplify.

$t=9+18$

$t=27$

So, the head is 9 inches long, the body is 36 inches long, and the tail is 27 inches long.

For the full length it is h+b+t, or $9+36+27$.

$9+36+27=72$

So your fish is 72 inches long (now that's a catch!)

Looks right to me, I do not understand what your question is. 

hi Juriemagic  

This is an unusual question :)

$5*3^x-9*3^{x-2} \over 3^x-3^{x-1}\\ =(5*3^x-9*\frac{3^x}{9})\div(3^x-\frac{3^x}{3})\\ =(5*3^x-3^x)\div(\frac{3*3^x-3^x}{3})\\ =(4*3^x)\div(\frac{2*3^x}{3})\\ =(4*3^x)\times(\frac{3}{2*3^x})\\ =(4)\times(\frac{3}{2})\\ =6$

What do you mean by 'The formula for a polygon is'

The formula for what???

thanx Alan,

this helps me tremendously...thank you..

First factor the three integers to see that:

$432=3*12^2\\675=3*15^2\\300=3*10^2\\\text{so:}\\\sqrt{432}=12\sqrt3\\\sqrt{675}=15\sqrt3\\\sqrt{300}=10\sqrt3$

You can take it from here.

I see, that makes sense...thank you kindly...

That is your final answer. It cannot be more simplified. 

With gamma and the beta function

Formula: $\displaystyle \int \limits_{0}^{\pi/2} \cos^{2u-1}(x)\cdot \sin^{2v-1}(x) \;dx = \frac12 \cdot B(u,v), \qquad \text{Re } u>0, \quad \text{Re } v>0$
Where $B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}, \qquad \text{Re } u>0, \quad \text{Re } v>0$

4)$\displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\;dx$

$\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 15 & 2v-1 = 0 \\ 2u = 16 & 2v = 1\\ u = 8 & v = \frac12 \\ \end{array} \end{array}$

$\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx &=& \frac12 \cdot B(8,\frac12) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \Gamma(\frac12)}{\Gamma(8+\frac12)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \Gamma(8) = 7! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \displaystyle \Gamma(8+\frac12) = \frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! 8! 4^8}{16!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 4^8}{ 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14\cdot 15\cdot 16} \\\\ &=& \displaystyle \dfrac{165150720 }{518918400 } \\\\ &=& \displaystyle \dfrac{2048\cdot 80640 }{6435\cdot 80640 } \\\\ &=& \displaystyle \dfrac{2048 }{6435 } \\ \end{array}$

5) $\displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx$

$\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 0 & 2v-1 = 13 \\ 2u = 1 & 2v = 14 \\ u = \frac12 & v = 7 \\ \end{array} \end{array}$

$\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx &=& \frac12 \cdot B(\frac12,7) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(\frac12) \cdot \Gamma(7)}{\Gamma(\frac12+7)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot \Gamma(7)}{\Gamma(7+\frac12)} \quad & | \quad \Gamma(7) = 6! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\Gamma(7+\frac12)} \quad & | \quad \displaystyle \Gamma(7+\frac12) = \frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{6! 7! 4^7}{14!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6 \cdot 4^7}{ 8\cdot 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14} \\\\ &=& \displaystyle \dfrac{5898240 }{17297280 } \\\\ &=& \displaystyle \dfrac{1024\cdot 5760 }{3003\cdot 5760 } \\\\ &=& \displaystyle \dfrac{1024 }{3003 } \\ \end{array}$

The number of bacteria  N minutes after 11:00   =  2^N

So when the bottle is full at noon , we will have 2^60 bacteria

And at 11:52 we will have 2^52 bacteria

So

2^52 / 2^60  =    1 / 2^8   full =   1 / 256  full at 11:52

[ A little surprising, huh ?? ]

a) No condition is imposed

We can use any of the seven letters in each position...so

7 * 7 * 7 * 7  =  2401 "words"

b) No letter can be repeated in a word.

We have 7 ways to pick the first letter, 6 ways to pick the second, 5 ways to pick the third and 4 ways to pick the fourth............so   7 * 6 * 5 * 4  =  840 "words"

c) Each word must begin with the letter A

If we can repeat letters, the other three positions can be filled in 7 * 7 * 7 =  343 ways....so..we have 343 "words"

If we cannot repeat letters we have   6 * 5 * 4  ways to fill the other three positions = 120 "words"

d) The letter C must be at the end.

Same answers as  (c)

e) The second letter must be a vowel.

If we can repeat letters....we have 7 ways to fill three of the positions and 2 ways to fill the remaining one [ since we have only two vowels ]...so....   7 * 2 * 7 * 7  = 686 "words"

If we cannot repeat letters, we have 7 * 2 * 6 * 5  =  420 "words"

Since AOB is 90°, the  whole area is just the area of 1/4 of a circle = pi (8)^2 / 4  =

64 pi / 4  =  16 pi  units^2       (1)

The area of the triangular part of the figure is(1/2) * 8^2  =  64/2  = 32 units^2     (2)

So...the purple area is just   (1)  - (2)  =   [16 pi - 32 ] units^2 

The other leg  = 5

And the area of the base is    (1/2) (5) (12)  =  (1/2) * 60  =  30 units^2

And the volume of the prism is

Base area * Height  =    30  * 10   =  300 units cubed

Let the smaller circle's radius  = r  ⇒ circumference =  2pi r

And the larger circle has a radius twice the smaller  = 2r  ⇒  circumerence = 2pi (2r)  =

4 pi r

So...the ratio of their circumferences  =

[ 2 pi r ] : [ 4 pi r ]  =  2 : 4  =  1 : 2

thx math wiz.