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Does such a polynomial exist?

We can isolate the x terms on one side.

201x=0

Then,

we divide by 201

x=0

So,

x=0

* l - large box

 s - small box

* l = s + 20

 * 2l = 1/2 (6s)

   2l = 3s

     l = 3/2s

* [s + 20 = 3/2s] 2

   2s + 40 = 3s

           40 = s

    *       l = 40 + 20

            l = 60 nails 

There are 2 cases to consider: 

Even - Odd - Even

Odd - Even - Even

For the first case, there are \(4 \times 5 \times 5 = 100\) numbers. (Remember, the leading digit can't be 0!!)

For the second case, there are \(5 \times 5 \times 5 = 125\) numbers. 

Thus, there are a total of \(100 + 125 = \color{brown}\boxed{225}\) numbers that meet these criteria.

Since O is the circumcenter, AO will be the circumradius (as will also be OB and OC).

Let X be the midpoint of AB.

Draw OX; OX will be perpendicular to AB.

Since AB = 10, AX = 5.

Triangle(AXO) will be a right triangle with right angle(AXO) and hypotenuse AO.

cos( angle(AX0) ) = 5 / AO

cos( 50^o ) = 5 / AO     --->     AO  = 5 / cos(50^o)

A right triangle with side lengths  6  and  b  has hypotenuse value  c.

Using the Pythagorean Theorem:  6² + b²  =  c².

The perimeter of the triangle is  24     --->     6 + b + c  =  24

                                                            --->                 b  =  18 - c

Substituting:                  6² + (18 - c)²  =  c²

                        36 + ( 324 - 36c + c² )  =  c²

                                     360 - 36c + c²  =  c²

                                             360 - 36c  =  0

                                                     -36c  =  -360

                                                           c  =  10

13  is the circumradius  = C

C  = 13  =    product  of the sides / [  4 Area ]

Let x =  BC

13  =           (22 * 22 * x)

             _________________________________________________

            sqrt  [  (22 + 22 + x ) ( 22 + 22 - x) ( x + 22 - 22) (x + 22 - 22) ]

13  =         [ 22 * 22 * x  ]

           ________________________

             sqrt  [ ( 44 + x)  ( 44 - x)  (x^2) ]  

13 =     [ 22 * 22 * x ] 

          ________________________

            x * sqrt [ (44 + x) * (44 - x) ]

sqrt [ (44 + x) (44 - x) ] =    484   /  13

sqrt [ 1936 - x^2 ]  =  484/13       square both sides

1936  -  x^2   =    (484)^2 / 169

x^2  =   1936   -  (484)^2 /  169

x^2  = 92928  / 169

x = sqrt [ 92928 / 169 ]  ≈  23.449  = BC

\($\sqrt{21}$, $\sqrt[3]{27}$, $-\sqrt{64}$, $\sqrt[4]{25}$, $\sqrt{40}$, $-\sqrt[3]{81}$ \)

sqrt (21)  =  (4. and some  decimal)

cube root  27  = 3

-sqrt (64)     = -8

4th root of 25  =  (5^2)^(1/4)  = 5^(1/2)  = sqrt 5  =  ( 2. and  some  decimal)  

sqrt 40  = 6.and some  decimal

-cube root of 81  =    ( -4. and some decimal )

You should be able to  do this, now

x (3x - 7 )   =  -1 + 4x

3x^2 - 7x  = -1  + 4x           simplify as

3x^2 - 11x + 1    = 0

By  the Q formula  we  haxe

solutions =       11 ± sqrt [ 11^2  - ( 4 * 3 * 1 ) ]                    11 ± sqrt [ 109 ]

                       _______________________  =                ______________

                                         2 * 3                                                     6

m = 11

n = 109

p = 6

m + n + p  =   126

Thank you!

Find BC.

Hello Guest!

\(f(x)=\sqrt{22^2-x^2}\\ g(x)=\sqrt{13^2-x^2}+13\\ \sqrt{22^2-x^2}=\sqrt{13^2-x^2} +13\\ 22^2-x^2=13^2-x^2+2\cdot 13\cdot \sqrt{13^2-x^2} +13^2\\ 13^2-x^2= (\frac{22^2-13^2-13^2}{26})^2 \\ x=\sqrt{13^2- (\frac{22^2-13^2-13^2}{26})^2}\)

\(x=11.725\\ \overline{BC}=2x\)

\(\overline{BC}=23.449\)

  !

Your base is  correct   !!!!

Draw a an altitude from the top right vertex to the base

This forms a 30-60-90  right triangle......which has sides of 1  , sqrt 3   and 2

sqrt 3   is the altitude  (height)

Can you take it  from  there ???

Note

1/a^2   +   1/b^2  =     [ a^2 + b^2 ] /  [ ab]^2

5x^2   -11x + 4   =  0

By Vieta

Sum of the roots =  11/5

a + b = 11/5       square both sides

a^2 + 2ab  + b^2   =  121/ 25 

a^2 + b^2  =  121/25 - 2ab      (1)

And product of the  roots  =  4/5

So

ab = 4/5    ⇒  [ab]^2  = 16/25

2ab = 8/5

So (1)  becomes a^2 + b^2 =  121/25 - 8/5  =   121/25 - 40 /25  =   81/25

So

1/a^2  + 1/b^2   =     [ a^2 + b^2]  / [  ab]^2   =   [ 81/25] / [ 16/25]  =   81/16

Whatever x  makes the denominator  = 0  is  not in the  domain

x - 64  =0 

64 is  not in the  domain

x^2 - 64  =  0     so.... ( x + 8) ( x -8)    = 0     ....so    x = -8  and  x = 8  are not in the domain

x^3 - 64  =  0

x^3 = 64

x = 4    is not in the  domain

x^4 - 64   =  0 

(x^2 - 8) (x^2 + 8)   = 0

x^2  - 8  =0  ⇒   x =  ±sqrt (8)

The second factor set to 0  produces a non-real solution 

So....not in the  domain  =  x =   -8,  -sqrt (8) , 4 , sqrt (8), 8 , 64

Using the quad formula:

\(\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}\)

\(\frac{\left(11\right)\pm \sqrt{41}}{10}\)

Substituting this into \(\frac{1}{a}^2\:+\:\frac{1}{b}^2.\)

\(\frac{1}{\frac{\left(11\right)\pm \:\sqrt{41}}{10}}^2\:+\:\frac{1}{\frac{\left(11\right)\pm \:\sqrt{41}}{10}}^2\)

= \(\frac{81\pm11\sqrt{41}}{16}\)

or 

approx. 9.46464646464 and 0.66035

-Vinculum

[150 km/h] / 60 minutes ==2.5 km/minute - speed of the train.

Length of the train + length of tunnel ==4 + 1/2 ==4 1/2 km - train must travel to emerge from the tunnel.

Time ==Distance covered / speed ==4.5 k / 2.5 km/m ==1.8 minutes - time it takes the train to go through the tunnel.

AB = BC =  20

Let the altitude  =  BM

And this altitude will bisect the base

Let (1/2) of the base =  AM

So  by the Pythagorean Theorem, the height  =    sqrt [ 20^2 - AM^2]

And

Area =  (1/2) base * height

160  = AM * sqrt  [ 20^2  - AM^2]           square both sides

25600 = AM^2  [ 400 - AM^2]

25600  = 400AM^2  - AM^4

AM^4  - 400AM^2 + 25600  =  0    factor  as

(AM^2 - 320) ( AM^2 - 80)  = 0

So  either

AM^2  -  320  =  0                    or               AM^2  - 80  =  0

AM^2  = 320                                              AM^2   = 80       

AM = sqrt (320)  =  8sqrt( 5 )                        AM =  sqrt (80) =  4sqrt (5)

We have two possible perimeters

20 + 20 + 2 (8sqrt 5)  =    40 + 16sqrt (5)   units

   or

20 + 20  + 2 ( 4 sqrt 5)  =   40 + 8 sqrt (5)    units

Write in slope intercept form: \(y = -{2 \over3}x+1\)

Add \({2 \over 3}x\) to both sides: \(y+{2 \over3}x=1\)

Multiply the equation by 3: \(\color{brown}\boxed{2x+3y=3}\)

We can simplify first:

x^2 + (b - 6)x + 14 = 0

Now we can use a bit of brute force here:

(x + p)(x + q) 

pq = 14

p + q = b - 6

(p, q) = (14, 1) (7, 2) (-14, -1) (-7, -2)

So we can plug the values of p and q in to get all the possible values of b:

14 + 1 = b - 6 --> b = 21

7 + 2 = b - 6 --> b = 15

-14 + -1 = b - 6 --> b = -9

-7 + -2 = b - 6 --> b = -3

Thus, all the possible values of b are: 21, 15, -9, -3.

⚡⚡⚡

ShockFish

The slope of the line is 2. 

The point \((16,c)\) is 11 units away from \((5,5)\)

Because the slope is 2, the line goes up 22 from 5, meaning \(c = 5+22=\color{brown}\boxed{27}\)

Chris will do everything BUT won't solve the equation... LOL!!!

Write as an equation: \((-2-x)^2+(0-y)^2=(0-x)^2+(2-y)^2\)

Setting \(y =0\), we have: \((-2-x)^2=(0-x)^2+4\)

Expanding, we get: \(x^2+4x+4=x^2+4\)

Solving, we get: \(\color{brown}\boxed{x = 0}\)

x^2 - 6x + 2 - 14  =  0

x^2  - 6x - 12   = 0

By Vieta, the sum of the roots  =  - (-6) / 1  = 6

So ...    r + s  = 6     square both sides

r^2 + 2rs + s^2  =   36     

r^2 + s^2  =  36 - 2rs (1)

And the product of the roots =  -12 /1 =  -12

So.....  rs = -12

So   2rs  = -24     (2)

So

( r - s)^2  =  r^2 - 2rs + s^2

 ( r -s)^2  = ( r^2 + s^2)  -  2rs         sub in (1)  and (2)

(r - s)^2  =  (36 - 2rs) - 2rs

(r - s)^2  =  (36 - -24) - (-24)

(r -s)^2  =  60 + 24  =   84