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sEVEN

noooooooooooooo

Hi hi hi hi hi hi hi

a)  With replacement   =  (5/12)^3  ≈   .072

b) Without replacement  = (5/12) (4/11) (3/10) ≈ .045

The product of two whole numbers is 96 and their sum is less than 30.what are the possibilities for the two numbers

\( xy=96\\x+y<30\)

4*24=96 4+24=28<30

6*16=96 6+16=22<30

8*12=96 8+12=20<30

  !

Volume  =  pi * radius^2 * height

Volume  =  pi * 4^2  * 11   =   176 pi cm^3

So......since the candles have the same vollume.....

176 pi   =  (1/3) pi * radius^2 * 20        divide both sides by (20/3) pi

[176/(20/3) ]   = radius^2

132/5   =  radius^2        take the square root of both sides

√ (132/5)   = raduis of conical candle ≈  5.14 cm

∛ t   -  1 / ∛t        we can write this as

t^(1/3)  -  t^(-1/3)       differentiating we have

(1/3) t^(-2/3)  +  (1/3)t^(-4/3)       factor out  (1/3)

(1/3) [ t^(-2/3) + t^(-4/3) ]  =

.333333 [ t^(-.666666) + t^(-1.33333) ] 

However.......we can actually express this more compactly as

(1/3)t^(-4/3) [ t^(2/3)  + 1 ]

Your statements contradict one another, as the commutative property of multiplication doesn't apply to this (which it should). Therefore, there is no real answer but you could either say that 3*4 is 12 or could say "Where is the question?".

10 cubed is 1000

like a boss

Multiply by √5 divided by √5:

\(\frac{2\sqrt3}{\sqrt5}\rightarrow \frac{2\sqrt3\sqrt5}{\sqrt5 \sqrt5}\rightarrow \frac{2\sqrt{15}}{5}\)

.

 4√48 + 3√147 + 5√12   =

4√ ( 16 * 3)  +  3√ (49 * 3)  +  5 √ (4 * 3)  =

4*4 √3  +   3*7 √3  +  5*2√3  =

16√3  + 21√3 + 10√3  =

47√3

Let the tree height  = h

And we have a rigrt triangle such that :

h^2  +  80^2   =  (2h + 11)^2     simplify

h^2  +  6400  =  4h^2 + 44h + 121

3h^2 + 44h  - 6279  = 0    

Solving this  using  the  quadratic formula and taking the positive answer  we get that h =39 ft

Here, h is the height of the tree.

from the Pythagorean theorem:

h² + 80² = (11 + 2h)²

h² + 6400 = (11 + 2h)(11 + 2h)

h² + 6400 = 121 + 44h + 4h²            Subtract h² and 6400 from both sides.

0 = -6279 + 44h + 3h²                      Rearrange.

0 = 3h² + 44h - 6279                        Use quadratic formula to solve for h.

\(h = {-44 \pm \sqrt{44^2-4(3)(-6279)} \over 2(3)} \\~\\ h = \frac{-44\pm278}{6} \\~\\ h=\frac{-44+278}{6}=39 \qquad\text{or}\qquad h=\frac{-44-278}{6}=-\frac{161}3\)

So...the height of the tree must be 39 feet

To  continue where ParthForThree left off

2x^2+6x-176=0     divide through by 2

x^2  +  3x  -  88  =  0        factor

(x + 11) ( x - 8)  = 0

Set both factors to 0  and solving for x  we get that x  = -11 ft  (reject)  or x  = 8 ft

This is the width    ......  and the length  =  2(8)  + 6  =  22 ft

Rewrite as 

-6 * x^(1/2) -   7 * x^(-8)

The dervative is

-3  x ^(-1/2)   +   56 x^(-9)  =

-3 / √x   +   56  / x⁹

The  sides are   5 - 12 - 13

Imagine folding this up like it says in the problem...we can see that

height of box = 5

width of box = w - 5 - 5 = w - 10

length of box = (w + 25) - 5 - 5 = w + 15

volume of box = (height)(width)(length)

750 = (5)(w - 10)(w + 15)

750/5 = (w - 10)(w + 15)

150 = w² + 5w - 150

0 = w² + 5w - 300

0 = (w + 20)(w - 15)

w = -20     or     w = 15

So...the original width must be 15 in.   

650000

Lw=area

x(2x+6)=176

2x^2+6x=176

2x^2+6x-176=0

What is this number extended into standard form.

2.28338e+33

\(2.28338\times 10^{33}\\\color{blue}= 2 \ 283 \ 380 \ 000 \ 000 \ 000 \ 000 \ 000 \ 000 \ 000 \ 000 \ 000 \)

\(2.28338 \ Quintilliarden\)

  !

Well you just made that too easy!  OK, I always get confused when I see the 1/2...thank you!  I'll be coming back for more.  

2x^1/2-9 = 11      add 9  to both sides

2x ^(1/2)  =  20      divide both sides by 2

x^(1/2)  =  10        square both sides

[ x ^(1/2) ] ^2   =  10^2

x  =   100

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