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Volume of cube = S^3, where S = one side

V = 16^3 =4,096 mm^3 - volume of the cube

Volume of square pyramid =S^2 x h/3, where S=side, h=height

V =S^2 x h/3

4,096 =16^2 x h/3

4,096 = 256 x h/3 Divide both sides by 256

16 = h/3 Cross multiply

h = 48 mm - Height of the pyramid

48 / 16 = 3x - higher is the square pyramid than the cube.

I think you meant B not D on 82 :P

You were right thanks so much!

I believe this will work, ty ^-^

Thank you so much, but my teacher says it's \(y=- \frac{x^2}{8}+2\)

Sorry, but can't read your LaTex!.

Thanx Tertre, I do appreciate...

I am terrible at this math stuff!

without brackets your meaning is not clear

This is your question

2√52x^9/√13x =  2(√52)x^9/(√13)x =  2(sqrt52)x^9*x/(sqrt13) 

But I suspect that you actually mean

\(\frac{2*\sqrt{52}\;x^9}{\sqrt{13}\;x}\\ =\frac{2*2\;x^9}{x^1}\\ =4x^8 \)

I do not think that this is the whole question.

Thanks!!!! 

Here's the first two , LF  :

https://web2.0calc.com/questions/please-help-asap-ty_5#r7

And here are the other  ones :

https://web2.0calc.com/questions/please-help-asap-tysm#r1

The final answer is actually, 512!! 

f(x)  = 2x  + 2[ x  ( 1 - x) ]^(1/2)

f'(x)   2x + 2 [ x - x^2]^(1/2)      take the derivative

f '(x) = 2 + [ x - x^2]^(-1/2) (1 - 2x)        set this to 0

2  = (2x - 1)  * [x - x^2]^(-1/2)

2 (x - x^2)^(1/2)  = 2x - 1        square both sides

4 (x - x^2)  =  4x^2 - 4x + 1

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

8x^2 - 8x + 1  =  0

8 (x^2 - x + 1/8)  = 0

x^2 - x + 1/8  =  0

x^2 - x  + 1/4  =  -1/8 + 1/4

(x - 1/2)^2  = 1/8        take the square root of both sides

x - 1/2 =  ±√(1/8)

x  = ±√(1/8) + 1/2

The positive root  gives the  x value of the maximum as  √[1/8] + 1/2  ≈ .3004

And the max value  is  ≈ 2.41421

\(x\cdot\lfloor x\rfloor=70 \)

Note that √70  ≈ 8.366

So... if we divide 70 by 8  we get  70 / 8  =  35 / 4 = 8 3/4  = 8.75

So.....8.75 * 8  = 70

And ....  the floor of 8.75  =  8

So

8.75 * floor [ 8.75 ] =

8.75  * 8    =

70

Then  x  = 8.75

ln ( x + 3)  - ln x  = 4

ln [ (x + 3) / x ]  = 4

e^ [ ln [ (x + 3 ) / x ] ]  = e^4        { remember that e^(ln a)  = a  }

[ x + 3 ] / x  = e^4      cross-multiply

x + 3  = xe^4   rearrange as

x - xe^4  =  - 3       factor out  x

x [ 1 - e^4]  = -3 

x  = - 3 / [ 1 - e^4]   

Ah, here it is!

The vertex has to be right in between them, so the vertex is \((0, 2)\). Now, we use the equation \((x-h)^2=4p(y-k)\). \(p = -2\) , because that is the directed distance between the vertex and focus. We get than \(x^2 = -8(y-2)\), so 

\(\boxed{y = -\frac{1}{8}x^2+2}\) is the desired equation.

√[25 - x^2 ] + √[ - ( x - 2) ]  =

√[25 - x^2 ]  + √ [2 - x ]

The domain for the first function  is  [ -5, 5 ]

The domain for the second function is  ( - infinity, 2]

Because  we must take the most restrictive interval, this is   [ -5, 2 ]

So....the width of the interval  is  7 units   [ 8 integers ]

So the graph, here : https://www.desmos.com/calculator/ioiruyyh4v

Because CE  is a chord meeting a tangent, the measure of the angle supplemental  to angle PCQ =

(1/2) ( m arc CDE)  =   (1/2) (m arc CD + m arc DE ) = (1/2) ( 70 + 30)  =  50°

So....angle  PCQ  =  180 - 50  =  130°

And angle  CQP is vertical to angle DQE  =  40°  

So....in triangle PCQ, angle PCQ  = 130  and angle CQP = 40

So  angle CPQ  =  180 - 130 - 40  =  10°

And angle  QPE  equals this  = 10°

And angle PQE is supplemental to  angle DQE  = 140°

So...in triangle PQE,  angle QEP  = 180 - 140 - 10  = 30°

So m arc CA  is twice this  = 60°

So   m arc AE  =  360 - m arc DE - m arc CD  - m arc CA  =

360  - 30 - 70 - 60  =

360  - 160  =

200°

This doesn't  "look " to be correct......[ I don't think I made any mistakes, but maybe I did !!!!]

a,Does that help