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Let the number of doublings=n, then:

2^n =1,000/100

2^n =10 take the log of both sides

n =3.3219 doublings in 5 weeks..

5/3.3219 =1.505 - weeks - period of 1 doubling, so that:

2^3.3219 =10x - population growth in 5 weeks.

We  have this function :

P(t)  =  100(e)^kt

And when t  = 5, we have that

1000 =  100(e)^k5            divide both sides by  100

10  =  e^5k        take the ln of both sides

ln 10  =  ln e^5k     and we can write

ln 10  =  5k  ln e                  {  ln e   =  1...so we can ignore this }

ln 10 =  5k        divide both sides by 5

ln 10 / 5  =  k

The function is

P(t)  = 100(e) ^{ln10 * t / 5}

So.....we want to find the doubling time....and we have that

200  = 100(e)^{ln10 * t / 5}       divide both sides by  100

2  =  e ^{ln 10 * t / 5}      take the ln  again

ln 2  = ln e ^{ln 10 *  / 5}      amd we can write

ln 2  =  [  (ln10) / 5 ] * t          

5 ln 2  =  ln 10 * t

5 ln 2 / ln 10   =  t  ≈     1.5 weeks

What Melody said  !!!!!

It is good that you are back :))

Thank you. :) you made it sound simple, haha, 

(-5)(-3)-(2)(3)

--------------------

3[5/(3-8]

Let's take the top, first

(-5)(-3)  - (2)(3)  =

15  -  6    = 

9       

The bottom is a little trickier

3 [ 5 / (3 - 8) ]  =         work inside the parentheses, first

3 [ 5 / (-5) ]       =       now, inside the brackets  

3 [  -1 ]    =           now, just multiply

-3

So.....we end up with

9  /  - 3    =

-3

Hello!! Been a long time. 

Ohh! Melody is here! :)

Hi Hayley   

-8 + 4- (-23) -13

Rememer that   a  "-"  outside any parentheses or brackets changes the sign(s)  on the term(s) inside....so we have.....

-8  +  4  + 23  -  13  =

- 4  + 23 - 13    =

-4  +  10

6

Welcome back, Hayley-Meister....things are a LITTLE different here...but mostly the same.....

Its been awhile. haha. Thanks!! :)

HAYLEY!!!!.....great to see you, girl.....I always wondered what became of you....sure.... post your questions  !!!!

5x^2  + 5x + 10

Factor out a 5

5 ( x^2  + x + 2)       {  the polynomial in the parentheses is irreducible }

-3 ( x + 3)  =

-3x  - 9

Heh yeah ask a seventh grader that(I am a seventh grader)

I love Bravery Sans!! 

Let the number of teachers=T, then:

12 / 1 =3,000 / T   Cross multiply

12T = 3,000          Divide both sides by 12

T =3,000 / 12

T = 250 Teachers

Thx 

Well it doesn't make sense to me, it says that division gives a polynomial g(x)

"Let f(x) = 3(x^4+x^3+x^2+1)/(x^2+x-2). Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity."

Was this the intended question?

Let f(x) = 3(x^4+x^3+x^2+1)/(x^2+x-2).

Find a polynomial g(x) such that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity.

Let's call the length of the class period  " p ".

From the problem statement, we can make this equation:

p  -  10%  of  p   =   45 minutes

p  -  0.1p   =   45 minutes

0.9p   =   45 minutes      (In other words, 90% of the class period is 45 minutes.)

                                       Divide both sides of the equation by  0.9 .

p   =   45 / 0.9 minutes

p   =   50 minutes          

The sum of the exterior  angles of a polygon  =  360°

So....the missing exterior  angle  is

360  -  255  =

105°

Wow! I actually understood it that way! Thanks so much, you're the bomb cphil.  Have a great day :)

I  =  V / R

Multiply both sides by R

R I   =  V     

Divide both sides by I

R  =   V  / I

Actually....this one is a liitle easier.....

If you can envision this

Let's call the vertex of the 73°  angle "A," and from here,  "travel"  up to the vertex  of  angle z

Let us call the vertex of this angle "B"

And from this vertex "travel'  up to the  vertex of the 131°  angle....let's call this vertex "C"

Now....from this vertex "travel"  over to the vertex of the 42°  angle

Let's call this vertex  "D"

Now....from this vertex, "travel"  back to "A"

So.....we have a quadrilateral ABCD

And the sum of the interior angles of a quadrilateral = 360°

So....this implies that

Measure of A  + Measure of B + Measure of C + Measure of D  =  360

So  we have

73  +  z  +  131  +  42   =  360       simplify

246  + z   =  360       subtract 246 from both sides

z  =  114°

Don't worry about asking too many questions.. sometimes..it's the only way to learn....!!!

Again...we have an irregular pentagon

Just like the other problem, the interior angles sum  to  540°

The "missing" angle in the pentagon  will be supplemental to A  and will =  180 - A

So.....we have that

115 + 96  +  118  + 104  +  ( 180 - A)  =  540       simplify 

613   -  A   =  540         add A to both sides, subtract 540 from both sides

73°  =  A