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THE equal number = x

25 x +    10 x  + 5x  = 120

40x= 120

x = 3  of each coin         Anne has   4 of each coin      =  12 coins

If you subtract the  'yes' on the R graph from the 'yes' on the L graph, you can envision that about 1/2 of the circle on the L still remains

   1/2  is  50 percent.....   this is a tricky worded question....but that is the answer I would submit.

Thanks!

Evaluate   floor ( - 23 / 9) = - 3

As follows:

This was my approach

\(y=\frac{(x^2 +14x+9)}{(x^2+2x+3)}\\ y=\frac{(x^2 +2x+3+12x+6)}{(x^2+2x+3)}\\ y=1+6*\frac{(2x+1)}{(x^2+2x+3)}\\ \text{I note that the denominator is positive for all real values of x}\\ \text{As x tends to }\pm\infty \text{ y tends to +1}\\ y'=6*\frac{(x^2+2x+3)(2)-(2x+2)(2x+1)}{(x^2+2x+3)^2}\\ y'=0 \quad when \quad \\ (x^2+2x+3)(2)-(2x+2)(2x+1)=0\\ 2x^2+4x+6-(4x^2+6x+2)=0\\ -2x^2-2x+4=0\\ x^2+x-2=0\\ (x+2)(x-1)=0\\ x=1\;\;or \;\;x=-2\\ \)

When x=1 y=4

When x=-2, y=-5

I now have enough info to know that the graph looks something like this:

\(\bf\text{So the range of this function is   }   -5 \le x\le 4\)

LaTex:

y=\frac{(x^2 +14x+9)}{(x^2+2x+3)}\\
y=\frac{(x^2 +2x+3+12x+6)}{(x^2+2x+3)}\\
y=1+6*\frac{(2x+1)}{(x^2+2x+3)}\\
\text{I note that the denominator is positive for all real values of x}\\
\text{As x tends to }\pm\infty \text{  y tends to +1}\\
y'=6*\frac{(x^2+2x+3)(2)-(2x+2)(2x+1)}{(x^2+2x+3)^2}\\
y'=0 \quad when \quad \\
(x^2+2x+3)(2)-(2x+2)(2x+1)=0\\
2x^2+4x+6-(4x^2+6x+2)=0\\
-2x^2-2x+4=0\\
x^2+x-2=0\\
(x+2)(x-1)=0\\
x=1\;\;or \;\;x=-2\\

There is no unique answer here, as you can see by factoring the number, which you can do here on the webcalc2.0 calculator:

6*5*3*67 = 6030

So you will need at least 6030 m^3 of storage.  But you will probably need more because the boxes have to fit without being crushed.

If they were in a straight snake that was 67 boxes long then it would be exactly 6030m^3

OR = √65  

WOW!

That is really cool !

question  x = ?

sinx +cosx = √6/2

Hello Guest!

\(sinx+cosx=\frac{1}{2}\sqrt{2}\\ sinx+\sqrt{1-sin^2x}=\frac{1}{2}\sqrt{2}\\ sinx=u\\ u+\sqrt{1-u^2}=\frac{1}{2}\sqrt{2}\\ \sqrt{1-u^2}=\frac{1}{2}\sqrt{2}-u\)        \(sinx-\sqrt{1-sin^2x}=\frac{1}{2}\sqrt{2}\)   (also pursue!)

\(1-u^2=\frac{1}{2}-\sqrt{2}\cdot u+u^2\\ 2u^2-\sqrt{2}\cdot u-\frac{1}{2}=0\\ u^2-\frac{1}{2}\sqrt{2}\cdot u-\frac{1}{4}=0 \)

\(u=\frac{1}{4}\sqrt{2}\pm \sqrt{\frac{1}{8}+\frac{1}{4}}\\ u=\frac{1}{4}\sqrt{2}\pm \sqrt{\frac{3}{8}}\\\)

\(u_1=sinx_1=0.96592\\ u_2=sinx_2=-0.25882\)

\(x_1=105°\\ x_2=-15°\)

  !

Jonathan and Victoria can finish a job in 13 days. Jonathan can do the job himself in 26 days. if Victoria wanted to do the job alone, how long would it take her?

\((j+v)\cdot 13days=1\\ 26days\cdot j=1\\ j=\frac{1}{26days}\)

\((\frac{1}{26days}+v)\cdot13days=1\\ v=\frac{1}{13days}-\frac{1}{26days}=\frac{26days-13days}{13\cdot26\ days^2}=\frac{13days}{13\cdot26\ days^2}\)

\(v=\frac{1}{26days}\)

Victoria can also do the job in 26 days.

  !

Thanks EP.  I will just add my 2c worth

The line appears to go through   (1,0)  and (-1,-1)

\(gradient=\frac{rise}{run}\\ gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\ gradient=\frac{y_2-y_1}{x_2-x_1}\\ gradient=\frac{-1-0}{-1-1}\\ gradient=\frac{-1}{-2}\\ gradient=\frac{1}{2}\\\)

now to get the equation I usually just think to myself 

 gradient =gradient

\(gradient=\frac{y_2-y_1}{x_2-x_1}\\ \text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\ \frac{1}{2}=\frac{y-0}{x-1}\\ 1(x-1)=2(y-0)\\ x-1=2y\\ x-2y-1=0 \)

(0,0) is in the wanted part of the graph so you have to sub in (0,0) and choose the correct inequality

(it is a dashed line, not a solid one, so there will be no equal sign)

\(LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\ LHS=0-0-1\\ LHS=-1\\ LHS<0\\ so\\ x-2y-1<0\)

LaTex:

gradient=\frac{rise}{run}\\
gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\
gradient=\frac{y_2-y_1}{x_2-x_1}\\
gradient=\frac{-1-0}{-1-1}\\
gradient=\frac{-1}{-2}\\
gradient=\frac{1}{2}\\

gradient=\frac{y_2-y_1}{x_2-x_1}\\
\text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\
\frac{1}{2}=\frac{y-0}{x-1}\\
1(x-1)=2(y-0)\\
x-1=2y\\
x-2y-1=0

LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\
LHS=0-0-1\\
LHS=-1\\
LHS<0\\
so\\
x-2y-1<0

I think this has been answered before...though I do not kow if I was correct.....this one is hard to read....

I believe the range could go as positive as   -5 x -2 = 10

   and as negative as -2 x 3 = -6

so  [-6,10]      Let me know if this is correct...or incorrect       (b would be '10' )

never mind, it doesn't show.

(2, 4, 6, 8, 10, 12, 16, 18, 20, 24, 30, 32, 36, 40, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 144, 160, 162, 180, 192, 216, 240, 270, 288, 320, 324, 360, 432, 480, 486, 540, 576, 648, 720, 810, 864, 960, 972, 1080, 1296, 1440, 1620, 1728, 1944, 2160, 2430, 2592, 2880, 3240, 3888, 4320, 4860, 5184, 6480, 7776, 8640, 9720, 12960, 15552, 19440, 25920, 38880, 77760)>>Total = 72 such divisors

The three angle summed = 180

  the two with variables  summed =  180 - 25 = 155

sooooo    17x-1    +    3x-4   = 155      solve  for x   then angle A = 3x-4

y = 3/x      to find inverse, switch x and y   then solve for y

x = 3/y

y = 3/x     is the inverse......     if you put the original function in there , you should get 'x' as a result

y = 3/ (3/x)   = x      

If you look at the graph of  3/x    you might be able to realize it is also the inverse:

https://www.desmos.com/calculator/pycz5hhztx

7      three angles of a triangle sum to 180    soa angle A = 60^o

              the second triangle    60 = 3x+15     solve for x

8     180   -   42  - 108  = 4x-4       solve for x   then compute 2x

(x-1) * f(x) = 3x^4+x^3-25x^2+38x-17        f(x) must be degree 3     after multiplication by 'x' in (x-1) the nuw fxn bumps up to 4th degree

leading term must be  3 x^3

constant must be  17      ( - 1  from the first part   *   17 from f(x) will give  -17 in the final )

divide    (x-1)   into   3x^4+x^3-25x^2+38x-17     to find  f(x)

ikr, lol...it made my computer super laggy 

Good work!

x! - (x-3)! = 23

x!    and     (x-3)!   have a difference of 23

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

x = 4      just by looking at it......not sure how to do it mathematically... 

x = 4

4! =24

(4 - 3)! =1

24 - 1 = 23

23 / 23 =1