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First, notice that the quadratics in the expression only have real roots when its discriminant is nonnegative. The biggest possible value for the constant term is 552 because when it is 553, there will be no real roots.

The product of the roots of a quadratic is equal to the constant term divided by the quadratic term. Since the quadratic term is always 1, the product of the roots of these quadratics will always be equal to the constant term.

Therefore, the problem is reduced to the product of 1*2*3*4*5*6*....*551*552.

That is equal to 552! (552 factorial).

That number is giant, so IDK if you want me to type it out.

Hello Guest!

We're kinda looking for the number of 7s in the numbers 1-109.

The amount of numbers divisable by 7 from 1-109 is 15 (7, 14, 21, 28, ..., 105). 

However, some numbers are divisable 49 which is made out of 2 7s, 7^2.

The amount of numbers divisable by 49 from 1-109 is 2 (49, 98). 

So there are 15 numbers divisable by 7, 2 of which are divisable by 49. 

There are 13 numbers divisable by 7 and not 49, and 2 number divisable by 49.

Thus, our answer is 13 + 2*2= 17.

A cool way to do this is to just keep dividing. 

Btw, floor is just the highest integer lower than the answer. For example floor(3.5) = 3. 

floor(109/7) = 15

floor(15/7) = 2

floor(2/7) = 0 

We stop dividing when our final answer is less than 1. 

Then, we just add everything, 15 + 2 = 17. 

This works because dividing by 7 two times is just like dividing by 49.

I hope this helped. :)))))

=^._.^=

I got that Pat drove 3 hours and Brandon drove the rest.

If Bradon drove the entire way, then he would have drove 575 miles. 

But for each hour Pat drives, that amount decreases by 10 miles.

To get our result, we subtract 545 from 575 to get 30 and divide by 10 to get that pat drove 3 miles

Checking our work, 3*90 is 270 and 2.75*100 is 275. 270+275 is 545.

Brandon and Pat drove from Edmonton to Saskatoon, a distance of 545km. Brandon drove the first part of the trip at an average speed of 100 km/h and Pat drove the remainder at an average speed of 90 km/h. If the trip took Brandon and Pat 5.75 h, determine the number of hours each drove. 

Hello Guest!

\(\color{blue}v_B\cdot t_B+v_P\cdot t_P=545\\ 100\cdot t_B+90\cdot t_P=545\\ \color{blue}t_B+t_P=5.75\\ t_B=5.75-t_P\)

\(100\cdot (5.75-t_P)+90\cdot t_P=545\\ 575-100t_P+90t_P=545\\ -10t_P=545-575\\ 10t_P=30\\ \color{blue}t_P=3\ h \)

\(t_B=5.75-t_P\\ t_B=5.75-3\)

\(t_B=2.75\ h\)

  !

Let the 4 schools be   A, B , C and  D

A plays  B, C and D

B plays  C, D    (they have already played A)

C plays D  ( they have already played A and B

D  has already played A, B and C

The players  from A  play a total of

(4 players from A ) ( 4 players* 3 schools) (3 games  each)  =  144 games

The players from B  play a total of

(4  players  from B) (4 players *2 schools)(3 games each )  =  96 games

The players from C  play a total of 

(4  players from C) ( 4  players * 1 school ) (3 games each)    = 48 games

144 + 96 +  48     =   

48 ( 3 + 2 + 1)  =

48 (6)  = 

288 games

I challenge you to show your working Cal.

A

           M

B                     C

The  midpoint of  the hypotenuse  will  be  the circumcenter of  a  right triangle

Therefore   AM =  MC   =  sqrt (117)

So....the  hypotenuse =   2sqrt (117)   =  sqrt (468)

Let AB  = x   so  BC  = 30 - x

And we have this equation

x^2  + (30 - x)^2     =  468        simplify

x^2  + x^2   - 60x  + 900  = 468

2x^2  - 60x  +  432  =   0

x^2 - 30x  +  216  =  0

(x - 12) ( x -18)   = 0

So     depending  upon your preference   either   AB =  12   and  BC    =18   or vice-versa

The area  =  (1/2)  (product of the legs)   =   (1/2) 216   =  108

Not sure about this.....but....

This must imply  that

(7/8) a  + (2/8)b  +  c   = 0

a(7/8)  + b (2/8)  +  c   =    0 

(x , y)   = (7/8 , 2/8)  =  (7/8, 1/4)  =  P

5^30  / 8

Notice  the pattern

5 mod  8   = 5

5^2  mod 8 =  1

5^3 mod 8  =  5

5^4 mod 8  = 1

So....it appears  that     5^(2n)   / 8       always produces a remainder of  1

So the formula to calculate the remainder is

 \(a^{p-1}\equiv1(mod \space p)\)

When a and p are relatively prime.

So we get

\(5^{23}\equiv1(mod\space 8)\)

From this we can get

\(1^{23}\times5^7\equiv78125\equiv125\equiv5(mod\space8)\)

So the remainder is 5

One root is 8    So  (x-8) is part....this results in:
(x-8)(4x+3)             = 4x² -29x-24

so  4x+3

4x^2  +  bx    - 24

Since 8 is a  root....performing some  synthetic division

8  [  4      b         -24        ]

             32       8b + 256

    ______________________

      4    b + 32       8b  +  232

So

8b   + 232     =   0

8b    =  - 232

b  =   -29

So

4x^2   -  29x   - 24   =    0

(4x  + 3) ( x - 8 )  =  0

The linear term is   (4x + 3)

Well simplify, 

4x+6+7x-9=12x-7-x+blank

add like terms, 

11x-3=11x-7+blank

-3=-7+blank, 

blank=4

Jill's speed  =   distance  / time

So    equating speeds

x^2  - 11x   -22   =   (x^2    - 3x  - 54)  / ( x - 9)

x^2  - 11x  - 22  =  ( x - 9) ( x + 6)  /  (x -9)

x^2  -11x  - 22  =  x +  6       subtract the right side ftom  the left and we  get that

x^2  - 12x - 28   = 0       factor as

(x - 14) ( x + 2)  =  0

Setting  both factors to 0  and solving  for x produces

x  =14       or  x   =   -2

When x   = 14    the speed   is    (14)^2 -11(14)  - 22  =  196 - 154 - 22  =    20 mph   (pretty fast)

When x  =  -2    the speed is   (-2)^2  - 11(-2)   - 22  =   4 + 22   -22   = 4 mph   ( this answer  is reasonable)

Hi, cryptoaops!

That diagram is ok, but my answer "Area of a triangle QRP = 3√3" is NOT correct!!!

Link:  https://web2.0calc.com/questions/geometry_49612

The correct answer is   [QRP] = 5.569219381 square units!!! 

Link:   https://web2.0calc.com/questions/quick-hard-question-help-asap

(I don't know how to convert that decimal into radical.)

AC  has  the slope  (-5/12)

And  the  equation of the  line through AC  is

y = (-5/12)x  +  5

12y   = -5x +  60

5x + 12y  - 60  =   0

The point  on AB  will be  (0, a)

And we  can solve this to  find "a"         (abs   =  absolute value )

abs (  5(0) +12a  - 60 )  / sqrt ( 5^2 + 13^2)   =  a

abs  ( 12a  -60_  / sqrt (169)

abs ( 12a - 60)   = sqrt (169)  a

abs  (12a  -60)   =13a

Either

12a   - 60  =  13a

-60  =  a        reject

or

60   -12a    =13a

60 =  25a

60/25   = a  =  12/5  = 2.4 = a   = the radius of the  circle

Here's a pic  :

sorry, yes it is all integers

Thanks!

Here we go, ABC is a right triangle, since AB=5 and BC=12, AC=13 due to Pythagorean theorem, the formula is 

area/semi perimeter = radius, the area is 30, then the semi perimeter is 15, 30/15=2

so the radius should be 2

It would be very helpful if you could present your strategy, my strategy was to use to variables, x as the length of the square and y as the length of GC and DH since they are equal

Can you show where you started?

y  = x   - 1/x

y =  [ x^2 -1 ] / x

xy   =  x^2    - 1

Trying to get x by itself  and then "swap" x  and  y is sticky

Here's one approach   by completing the  square on  x

yx  =  x^2  - 1          rearrange as

x^2  - yx     - 1    =   0       complete the square

Take 1/2  of y  =  y/2   square it  = y^2/4   add to both  sides

x^2  - yx   + (y^2/4)   =  1  + (y^2/4)       the left side  factors as

( x - y/2)^2  = ( 4 + y^2) / 4

x - y/2  =  ± √(4 + y^2)  / 2

x =  ± √(4 + y^2)  /  2  + y/2

x  =   [ y ± √ (4 + y^2) ]   / 2      "swap"  x and  y

y =  [ x ± √ (4 + x^2 ) ]  / 2   =    the "inverse"

Note  that  if  x  =  1   in the  first function, then  y   =  0

So   the point    ( 0, 1)   should  be  on the  "inverse" graph...so

1  =  [ 0 ± √ [ 4 + 0  ]   ] / 2

1  = ± √4/2

1  =  ±1

This is  only  true  if we  take the positive root

So.....  x  - 1/x     has no inverse   on   (-inf, 0)  U (0, inf)  because  we don't get a unique point for the "inverse"

However....if we restrict the domain of  g (x) to    (0, inf)    then  the inverse  becomes

y =  [ x +  √ [ 4  + x^2 ]   ]  /  2

Now.....note if   x =   2  in  g(x)   then   g(2)  =  3/2

And the point  ( 2,3/2) is  on g

So....the point   (3/2 , 2)   should be  on the  inverse   of  g

So

2 =   [ 3/2 +  √ [ 4 + (3/2)^2 ]   ]/  2

4 =  3/2  +    √ [ 4 + 1.5^2 ]   

4 =  3/2  +  √ [ 4 + 2.25  ]   

5/2  =  √(6.25  

2.5   = 2.5      which is true

So we  have an inverse if we  restrict the  domain  to  (0, inf)

Also.....look at the  graph here  for  x  -1/x

https://www.desmos.com/calculator/g4o70vl223

This  graph  does not pass the horizontal line test  ( a requirement for an inverse)

However....if we  restrict the  domain  to   (0, inf)   we will have a  graph that will pass the horizontal line test,  and so, it will have an inverse

23.5 miles would be least.....rounds to 24 miles