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Doesn't integrating that simply give us \(f(x) = c\)?

In that case, "all possible values of \(f'(0)\)  are simply... \(0\).

Thank you!

1 - 1/4 - 5/16 -3/8 = D

1 - 4/16 - 5/16 - 6/16 = D = ....................

I'd love to interact!

I'm not quite sure what the answers are... How would we continue to integrate f(x) = Cf'(x) to achieve the possible values of f'(0)?

Thank you both and happy holidays!

Sorry, but this is wrong

The answer is 1296.

y = 16/7.

$\sin (\angle BAD) = \frac{2 \pi}{3}$.

Ratio

g: p

2/6  =  g/24    solve for g

(2/6) * ( 24) = g = 8 glasses

(3 + x)/(5 + x)    =    (1 + x)/(4 + x)       cross multiply

(x+3)(x+4)   = ( x+1)(x+5)                   expand

x^2 + 7x + 12   =   x^2 +6x + 5            simplify

x = - 7

If roots are 2 and 4 then

(x-2)(x-4) = 0    expand

x^2 -6x + 8 = 0      when x = 8    this   equals  24  

       it needs to be = 6   

             so multiply the whole thing by 1/4   ( this will not change the zeroes)

1/4 x^2 - 1.5 x + 2 = 0  

sin a = 1/4   then cos a = sqrt(15)/4

sin 2α = 2 sin α cos α         sin and cos are both positive in Q I   (from 0 to pi/2  is Q I )

  sin 2a = 2 (1/4) (sqrt(15)/4)     <==== take it from here !  

pen = 5         notebook = 7

Just to add that "d" in the above stands for the difference between terms in the arithmetic sequence.  

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dont you understand? i said "STEPS"

1)    25%

2)    12.5%

3)    37.5%

4)    60%

y + 6  + d  = 12

   so   d = 6-y

12 + d = 3y

 d = 3y-12

equate the d's    (red equations)

6-y = 3y-12

4y = 18

 y = 4.5

The first term of an arithmetic progression is  4.

The common difference is  d.

The fifth term is  4 + 4d.

The sum of the first five terms is:  Sum  =  5[ 4 + (4 + 4d) ] / 2  =  20 + 10d

The sixth term is  4 + 5d

The tenth term is  4 + 9d

The sum of the next five terms is:  Sum  =  5[ (4 + 5d) + (4 + 9d) ] / 2  =  20 + 35d

Since the sum of the first five terms is one-fourth the sum of the next five terms:

              20 + 10d  =  ¼( 20 + 35d)

Solving:  d  =  -12,'

Since you know the first term and the common difference, you can find the 15th term.

When  x = 0,  y = 8    --->   the height of the triangular region is 8.

When  y = 0,  x = 4    --->   the width of the trianglular region is 4.

Use the formula for the area of a triangle:  Area  =  ½ · width · height  to get the area of the region.

I just realized that this question is an AoPS question.  It is against their rules for you to post here,  

they want you to use their notice board and your own brain.

Sorry, We cannot help you.

The base of a solid is the area between the parabolas \(x=y^2 \) and \(2y^2 = 3 - x \) . Calculate hte volume of the solid given that the cross sections perpendicular to the -axis are equilateral triangles!

This is how I see it.

If  the y value at  a given point is k then the side length of the equilateral triangle will be 2k and the vertical height will be (sqrt3)k

The area of the triangle (the cross section) would be 0.5*2k*(sqrt3)k = (sqrt3)k^2

\(\displaystyle \int_0^1 \sqrt3*(\sqrt x)^2\;dx\;+\;\;\int_1 ^3 \sqrt3 *\left[\sqrt{\frac{3-x}{2}}\right]^2\;\;dx\\ =\displaystyle \int_0^1 \sqrt3*x\;dx\;+\;\;\int_1 ^3 \sqrt3 *\;\frac{3-x}{2}\;\;dx\\ =\frac{\sqrt 3}{2} \left[\displaystyle \int_0^1 2x\;dx\;+\;\;\int_1 ^3 \;(3-x)\;\;dx\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle \left[x^2 \right ]_0^1+\;\;\left[\;(3x-\frac{x^2}{2})\right]_1^3\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle1+\;\;\;(9-\frac{9}{2})-(3-\frac{1}{2})\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle1+\;\;\;(\frac{9}{2})-(\frac{5}{2})\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle3 \right]\\~\\ =\frac{3\sqrt3}{2}\;\;\;u^3 \)

Glasses to plates ==2 : 6

24 / 6 ==4 - a constant

4  x  2 ==8 - number of glasses.

So that:  8 : 24 ==2 : 6

Square side = x

BC = 6        AB = 10

10/6(6 - x) = 6/10(10 - x)

x = 3.75

t = e + 52     or     e = t-52

t+10 =  2(e+10)

t+ 10 = 2 ( t-52 + 10)    <===== guest made a math error from here to the next step

t + 10 = 2t - 104 + 20

-t = -74

t = 74  y/o  

Brand A = 2/5 of mobile 

Brand B = 4/5 of remaining

Brand C = Rest

After shipping

Brand A = 20% = 1/5

Brand B = 25% = 1/4

Brand C = 1/3

Left mobiles = 228

Let total phones are x

Brand A = 2/5x

Left = x - 2/5x = 3/5x

B = 4/5 * Left = 4/5 * 3/5x = 12/25x

C = x - (2/5x + 12/25x) = 3x/25

After shipping out

A = 1/5 * 2/5x = 2/25

B = 1/4 * 12/25x = 3/25x

C = 1/3 * 3x/25 = 1/25x

Total shipped = 2x/25 + 3/25x + 1/25x

Total shipped = 6/25x

Left = 228

x - 6/25x = 228

    19/25x = 228

      x = (228¹² * 25)/ 19

      x = 300

a) fraction shipped out = Shipped/Total

= 6/25x/300 = 6/25 = 6/25 

b) A left in warehouse = 2/5x - 2/25x

                                    = 120 - 24

                                    = 96 
    C left in warehouse = 3x/25 - 1/25x

                                    = 2/25 * 300

                                    = 24

A are more than C = 96 - 24

                              = 72