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c = original amount of chocs

3/4 c   = dark    1/4 c = w

(1/4 c ) 1/2 = 3/4 c - 90                     c = 144

(3 ÷ 8)   x  100% = 37.5 %

x = origianl amount     

x - 1/5x + 5     after d      =  4/5 x + 5   remaining

(4/5x+5)  - (2/3 ( 4/5 x + 5) +3 )    =  4/15 x - 4/3   remaining after A

then Ben takes 1/4   leaving 3/4         3/4 (4/15 x - 4/3)  = 21

                                                                           x =  originally 110

Hello RunningMan!

No i wrote it. I was looking at them and they both had same words.

i wonder if anyone is going to answer

10c5   10c6  10c7   10c8   10c9  10c10

252 +   210 + 120  + 45 +   10     + 1  =                   ( I think !)

(a)    Find the number of toy planes in Box A in the end.

(b)    In the end, how many more toy cars were there in Box B than Box A?

Hello RunningMan!

     A                                             B

12c+20p                                 28c+40p                     beginning

12c+20p+xc -yp                     28c+40p-xc+yp          

(12+x)c+(20-y)p                     (28-x)c+(40+y)p

(12+x):(20-y)=4:3                   (28-x):(40+y)=1:2        end

36+3x=80-4y                          56-2x=40+y               

\(y=-9-0.75x+20\\ y=16-2x\\ 16-2x=11-0.75x\\ -1.25x=-5\)

\(x=4\\ y=8\)                                         \(x=4\\ y=8\)

\((12+x)c+(20-y)p\\ (12+4)c+(20-8)p\\ \color{blue}16\ cars+12\ planes\)           \((28-x)c+(40+y)p\\ (28-4)c+(40+8)p\\ \color{blue}24\ cars+48\ planes\)

\(24\ cars-16\ cars=8\ cars\)

\((a)\ The\ number\ of\ {\color{blue}toy\ planes\ in\ Box\ A}\ in\ the\ end\ is\color{blue}\ 12. \)

\((b)\ {\color{blue}8\ cars\ more}\ toy\ cars\ were\ there\ \color{blue}in\ Box\ B\ than\ Box\ A.\)

  !

Here's Phill's graph  (https://web2.0calc.com/questions/help_37192)

If AM = BM then these 2 arcs have the same length.

AM = √50        BM = √50

29/289*z < 16/161*z - 16

subtract 16/161  z from both sides 

   then you'll divide by the coefficient of z

  45/ 46259  z <    - 16

        z < -16 (46259/45) 

(x-1-sqrt2)(x-1+sqrt2)(x-2-sqrt2)(x-2+sqrt2) =

x^4- 6x^3+ 9x^2-2        after all of the multiplication   ( I used an online calculator)

it's actually 4

I would like to see someone do this one too.    

The side lengths of a rectangle have been measured to the nearest half of a meter as 7.5 and 18.5.

What is the greatest possible percent error in finding the area of the rectangle?

the smallest possible sides could be  7 by 18  area = 126      difference from estimate = 12.75      12.75/126 = approx 10.1%

The estimate area = 7.5*18.5 = 138.75

the longest possible sides could be 8 bu 19    area = 152      difference from estimate = 13.25        13.25/152= approx 8.7%

I am no statistician, there is probably some more refined way to do this but I would assume the answer is 10.1% (to 1 dec place)

Seems I made that more difficult than necessary.....

rotate the vector 90 degree counter clockwise

cos 90     - sin 90

sin 90       cos 90        or :           0     -1

                                                   1      0

-8r<-12 + 2r.     subtract 2 r from both sides

-10r < -12        divide both sides by -10 ....note: this flips the <  to >

r> 12/10

r > 6/5

120 = -6t^2 + 51 t

-6t^2 + 51 t -120 = 0    Quadratic formula show t = imaginary numbers ..... no answer ...as asinus found !

In how many seconds will the projectile first reach 120 feet in height?

Hello Guest!

\(f(t)=-6t^2+51t\\ \frac{df(t)}{dt}=-12t+51=0\\ t=4.25\)

\(h_{max}=-6\cdot 4.25^2+51\cdot 4.25\)

\(h_{max}=\color{blue}108.375\ feet\)

The projectile cannot reach a height of 120 feet.

  !

Thanks

But again I ask.

How did you determine this?

I made an error... 

CM = √72    or   3√8

I guess... nobody's perfect. 

A while ago I've seen Phill's graph.

So, CM happens to be √32   (2√8)

Neat job, Melody!

Given a right triangle ABC with altitude BD drawn to hypotenuse AC. If AD=4 and DC=8, what is the length of BD?

BD/4 = 8/BD

BD = sqrt(4 * 8)

BD = 2√8

Annie has attempted 21 free throws.

r < 3/7.

There are 245 numbers with units digit 0 or 1.

Angle B = 58 degrees.