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Thank you both!!

Yah, that is a mistake....it SHOULD be :

(82+93+68+93+70+98 +?) / 7 = 85

(  504 +? ) = 595

? = 91

Sorry, EP.....didn't see you doing this one, too !!!!

2x + 3y = 15.5       (1) 

 x  +  y =   6     ⇒   y = 6 - x     (2)

Sub (2) into (1)

2x + 3(6  - x)  =  15.5

2x + 18 - 3x = 15.5

-1x + 18 = 15.5               subtract  18 from both sides

-1x = -2.5   

x = 2.5

And y =   6 - 2.5 =  3.5

So    ( x , y)  = ( 2.5, 3.5)   

2x + 3y = 15.5

x+y = 6        or    x = 6-y    Substitute THIS value of x into the first equation

2 (6-y) + 3y = 15.5

12-2y+3y = 15.5

y = 3.5                          Now sustitue THIS value of y in to the first OR the second equation to find x

x + 3.5 = 6

x = 2.5

Let (1/x) be the amount of the pool filled  by a large pump in 1 minute

Let (1/y) be the amount of the pool filled by a small pump in one minute

So we have this system

4(1/x)(120) + 2(1/y)(120) =  1   ⇒  480/x + 240/ y = 1     (1)

2(1/x)(120) + 6(1/y)(120) = 1   ⇒   240/ x + 720/y = 1    ⇒  -480/x - 1440/ y = -2   (2)

Add (1) and (2)

-1200/ y = - 1    ⇒  y = 1200

And   

480/x + 240/1200 = 1

480/x + 1/5 = 1

480/x = 4/5

x/480 = 5/4

x = 480 (5/4) = 600

So.....the large pump fills 1/600 of the pool in a minute and the small pump fills 1/1200 pool every minute

So....in each minute they filll  1/400 of the pool working together

So.....8 times this = 8/400 of the pool  = 1/50 of the pool

So.....  (1/50) * minutes = 1/2

Minutes = 50/2 =   25 minutes

ceiling  (110/6)     =  19  packs 

LOL!!!!.....sometimes I make lucky guesses....!!!!

 V(h) = h(–h + 10)(–h + 8)  =  h ( h^2 - 18h + 80)  = h*3 - 18h^2 + 80h

Look at the graph here, LGB...

The max volume on the interval  occurs at (2.945, 105.028)

So...the max volume is about 105 ft^3

Thanks CPhill! You got it right! 

Thank you but is it possible to show me how you have arrived at that answer? 

One scenario could be as follows:

43^2 - 11^2 + 5^2 - 2^2 = 1749

Here is another way:

59^2 - 43^2 + 11^2 - 2^2 = 1749 {This does not meet the condition: a>3b>6c>12d}

Note: Simply by trial and error.

-3 * e^(2m - 5)  - 7 = -34  

Add 7 to both sides

-3 * e^(2m - 5) = -27

Divide both sides by -3

e^(2m - 5)  =  9         tale the Ln of both sides

Ln e ^( 2m - 5)  = Ln (9)        and we can write

(2m - 5) * Ln e  =   Ln (9)      since Ln e = 1,  we can ignore this

2m - 5   = Ln (9)       add  5 to both sides

2m =  Ln (9) + 5        divide both sides by 2

m = [ Ln (9) + 5 ] / 2  ≈   3.5986

It was right thank you!!

You're welcome.....!!!!

x^2 + x - 6 = 0        factor

(x + 3) ( x - 2) + 0

So

x + 3 = 0              or         x -  2 = 0

x = - 3                              x   =  2

oh wow that's a lot. thank you

Not too familiar with this....but....I believe that you are correct....

 (a+bi)^2 = 3+4i

(a + bi)^2  =   a^2 + 2abi - b^2

Equating coefficients, we have

a^2 - b^2  =  3     (1)

2ab = 4    ⇒  ab = 2  ⇒  b = 2/a    (2)

Sub (2) into (1)

a^2 - (2/a)^2 = 4

a^4 - 4  = 3a^2

a^4 - 3a^2 - 4 = 0        factor

(a^2 - 4) (a^2 + 1) = 0

A real  solultion will come from

(a^2 - 4) = 0

(a + 2) (a - 2) = 0

Since a is positve, then a = 2

And b = 2/a = 2/2 = 1

a + bi   =     2 + i

Going in reverse order

Note that the largest sum possible is

9, 8, 7, 6 , 5    =  35

So...there won't be too many of these !!!

9, 8, 7, 6, 3 =  33

9, 8, 7, 5, 4 =  33

No other sums = 33  utilizing 9, 8, 7 are possible

The next largest possible sum is

9,8,6,5,4 =  32

So....the only two lists I find are

3,6,7,8,9      and    4,5,7,8,9

It all depends whether the $900 interest was "simple interest" or " compounded interest"
1 - Simple interest:
Let the amount she invested 8 years =A
A x 0.08 x 8 years =$900, solve for A
A = $1,406.25 - what she invested 8 years ago at "simple interest"

2- Compounded interest
FV =PV + interest
A +900 =A x  {1.08}^8, solve for A
A =1,057.67 - This is what she invested if $900 was "compounded."

f(x)  = 1 / x^4

Notice that  the unction is not defined at x = 0

But as x gets very close to 0  .......the numerator of the fraction is much greater than the denominator......and the range goes to infinity

As x moves out both ways from 0 to some large neg / pos   x values....the denominator is much larger than the  numerator........and the y value of the function approaches 0  in both directions

So....the range is (0, inf )

thax

F(x)  = 2x^3  + x^2 - 11x

Don't know if you have had Calculus.....but we can find these quickly 

Take the derivative and set to 0

F' (x)  = 6x^2 + 2x - 11  = 0

The solutions to this are x = -1.53   and x = 1.2

The second derivative is

12x + 2

Plugging   the first x value into this results in a negative.....so we have a relative max at x = -1.53

And plugging the second value into this results in a positive.....so we have a relative min at x =1.2

Putting (-1.53) into the original function results in y = 12.01

Putiing (1.2) into the original function results in y = -8.3

So.....the rel max is (-1.53, 12.01)   and a rel min at (1.2, -8.3)

So....the second answer is correct