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Well, first multiply the nuber of boxes times the number of books. (5x8=40)

Now, substract 40 by 3 (40-3=37)

So, 37 was Isaac's reading goal.

:)

Hope that helped!

-SomeRandomGuyInTheInternet

It wouldn't be negative; i did this before, it's 0.13. Hope I helped, please pay forward and answer mine as well if you can :)

oh i though B oops

I"d say  " Personal Interview "

Good job,  creepercraft   !!!!!

Thank you!

C ( 12,5)  *  C(9.3) * C(6, 4)   = 997920 ways

I think you mean "vertex" form

y  =  x^2 + 5x +  36

Take 1/2 of 5  = 5/2...square this  = 25/4.....add it and subtract it on the right

y =  x^2 + 5x  + 25/4  + 36  - 25/4     factor  and simplify

y = ( x + 5/2)^2  + [ 144 - 25] / 4

y = ( x + 5/2)^2  + 119/4   ⇒   vertex form

The vertex  is  (-5/2, 119/4)

Let the vertex of the angle lie at  (0,0)

Note that  we move 6  units to the right and 4  units up to get to the point  ( 6 ,4)  that is the top right vertex of the triangle

So...we have the points  (0,0)  and ( 6,4)

And  slope  =  [  subtract the y  coordinates of each point in some order ]

                       __________________________________________________

                        [ subtract the   x coordinates of each point in the same order ]

So we have

[  4 - 0 ]               4               2          rise

_______   =       ___   =      __   =   ____

[ 6 - 0  ]                6              3          run

f(x)  = (x + 6)^2  for x  ≥ 6

For  f(x)......write y

y = ( x + 6)^2      take the square root of both sides

√y =  x + 6     subtract 6 from both sides and "swap"  x and y

√x  - 6   =   y      for y, write g(x)

So

g(x)  =  √x - 6   

Since  the first point on the original graph  is  (-6,0)

The first point on this graph  is  (0,-6)

So    x ≥ 0

Solve for x:
sqrt(x + 5) = sqrt(x) + 1

Raise both sides to the power of two:
x + 5 = (sqrt(x) + 1)^2

Subtract (sqrt(x) + 1)^2 from both sides:
5 - (sqrt(x) + 1)^2 + x = 0

5 - (sqrt(x) + 1)^2 + x = 4 - 2 sqrt(x):
4 - 2 sqrt(x) = 0

Subtract 4 from both sides:
-2 sqrt(x) = -4

Divide both sides by -2:
sqrt(x) = 2

Raise both sides to the power of two:
x = 4

This is how it looks would it still be no soultion ? i am kinda confused.

the square root goes only goes across x+5 and the square root just goes on x for the second equation

As you have it written....no  solution

Think about this....whatever √ x  is on one side it is the same thing on the other side

But  adding 5 to this value on one side will always be  3 more than adding 2 to it on the other side

So.... the left side > right side for any  x value  ≥ 0

I got it right thank you again! I do get confused on the solving but your explanation has helped .

I posted another question a bit different then this one , however I was having more diffucly with it , you could explain it? It is named "Solve for X" .

The inverse reverses the original coordinates...so.....

(1, -1)     (0, 0)  and (4,2)  are on the inverse graph

can you check number 6

Let's see if I remember this....

A function has to pass the "horizontal line" test to be a strict inverse....that is......any horizontal line drawn can only cut the function one time....

So...the two functions on the left will be strictly invertible, but not those on the right  

4x^3−12x^2 /4x^2+7x−2/ 2x^2−6x / 5x^2+11x+2

Invert the second fraction and multiply by the first

4x^3 - 12x^2               5x^2 + 11x + 2

___________     x     _______________                factor as much as possible

4x^2 + 7x - 2                 2x^2 - 6x

4x^2  ( x - 3)                    ( 5x + 1) ( x + 2)

______________   x      _______________       "cancel"  where possible

 (4x - 1) ( x + 2)                 2x ( x - 3)

2x * ( 5x + 1)

_____________

  ( 4x - 1)

Yes! Thank You, thats the way I solved it as well .If you could check the other one by chance ?

Invert    [ 3 + x ] / x    and multiply by the first fration

[ 2 + x ] / [ 8x ]  *   [x] / [ 3 + x ]  =

[ 2 + x ] [ x ]

____________          cancel  the x's

[ 8x ]  [ 3 + x ]

  2 + x

_________

8 ( 3 + x)

  Yep....good job, Jenny  !!!

I'm sorry that I don't know how to solve such a modular equation, with such large integers, in any formal manner. Maybe heureka or somebody else can provide such a solution.

However, I can easily write a very short computer code to search for the smallest positive n. Just in case you or your teacher are interested, here is the computer code that searched from 1 to 2,000,000 numbers in a matter of seconds and came up with: 1,234,577 as the smallest positive n.

{a=23^11; b=2000000; c=(1); n=2; if((a mod n)==1189872,c=(c,n),0); n++; if(n<=b,gotor-2,c)}

Check: 23^11 mod 1,234,577 =1,189,872. Good luck to you.

P.S. There are 2 larger numbers that also satisfy the equation:( 6,172,885, 8,123,897)

I think that you are correct, NSS !!!

I let you in on a secret, RP....

In most of the higher math classes I've had....I've rarely  encountered these types of radicals.....these are mostly " textbook" exercises.....once you have to suffer through this.....you probably won't see this stuff, again  !!!!!

[ Something to look  forward to, huh ???  ]