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how?

(2x^3y)^6 = 2^6.x^18.y^6

((2x)^3y)^6 = 2^18.x^18.y^6

2(x^3y)^6 = 2.x^18.y^6

omg dude XDD

I've done a few Monte-Carlo simulations of this and get 48 as the smallest value for N.  Here are a few arrangements that give this (remember that the numbers are arranged in a circle, rather than the straight line shown here):

16    22    10     7    28    13     1    19    25     4 

7    10    25    13     4    28    16     1    19    22

4    16    10    22     7    13    28     1    19    25

Welcome aboard matey! 

What about it?

Obviously I cannot see how your graph interacts with you.

But here is the graph of the first region. Perhaps from mine you can work out how to move the points (and what you are doing it for)

(x^2+y^2)^2+2x^2y^2-y^4=x^2(x^2+4y^2)

$(x^2+y^2)^2+2x^2y^2-y^4=x^2(x^2+4y^2)\\ $

$LHS\\=x^4+y^4+2x^2y^2+2x^2y^2-y^4\\ =x^4+4x^2y^2\\ =x^2(x^2+4y^2)\\ =RHS\\~\\ QED$

I have no idea how to do this but I would be interested if anyone else has an approach.   

The ratios of the measures of the sides of a triangle is 2:10:7.
Find the length of each side if the perimeter of the triangle is 228 meters.

$\begin{array}{|rclcl|} \hline a+b+c &=& 228 \quad & | & \quad \dfrac{c}{a} = \dfrac{7}{2} \Rightarrow \boxed{c=\dfrac{7}{2}a } \\ & & \quad & | & \quad \dfrac{a}{b} = \dfrac{2}{10} \Rightarrow \boxed{b=\dfrac{10}{2}a } \\ a+\dfrac{10}{2}a+\dfrac{7}{2}a &=& 228 \\ \dfrac{19}{2}a &=& 228 \\ a &=& \dfrac{228\cdot 2}{19} \\ \mathbf{a} & \mathbf{=} & \mathbf{24} \\\\ 24+b+c &=& 228 \\ b+c &=& 228-24 \\ b+c &=& 204 \quad & | & \quad \dfrac{b}{c} = \dfrac{10}{7} \Rightarrow \boxed{b=\dfrac{10}{7}c } \\ \dfrac{10}{7}c+c &=& 204 \\ \dfrac{17}{7}c &=& 204 \\ c &=& \dfrac{204\cdot 7}{17} \\ \mathbf{c} & \mathbf{=} & \mathbf{84} \\\\ b &=& \dfrac{10}{2}a\\ b &=& 5a\\ b &=& 5\cdot 24 \\ \mathbf{b} & \mathbf{=} & \mathbf{120} \\ \hline \end{array}$

Hi Dana :)

Welcome aboard, dana.....!!!!!

"Of " means "Multiply"

So

2/3 x 3/7  =   6 / 21    =  2 / 7

We have   2 + 10 + 7 =   19 equal parts

So

One side  =  (2/19) * 228  =   24

The next side is 10 : 2 =   5 times this =  120

So...the remaining side  =  228 - 120 - 24   =  84

f(2)  =  12

The inverse to f  =    (x / 3)^(1/2)  = f ^-1

The derivative of this is    (1/3)(1/2)(x/3)^(-1/2)  = (1/6)(x /3)^(-1/2)

So   

df^-1 / dx     at     f(2)   =  12   is

(1/6) (12/3)^(-1/2)  =

(1/6) ( 4)^(-1/2)  =

(1/6) / √4 =

(1 ) / ( 6 * 2)  =

1 /12

$z^5 = 1\\ z = \Large e^{i 2\pi \frac{k}{5}}, \normalsize k=0,1,\dots 4$

$\text{It might be a bit easier to visualize the quadrants if we write these as degrees}\\ r = (0, 72, 144, 216, 288)^\circ\\ \text{and clearly }180 < 216 < 270\\ 216^\circ \Rightarrow e^{i 2\pi \frac 3 5}\\ r = 1,~\theta = \dfrac{6\pi}{5}$

$GCF(42,70)=7\\ \text{Thus the common roots of unity are}\\ \Large e^{\pm i 2\pi \frac{k}{7}},~\normalsize k=0,1,\dots ,6$

$\text{There are 14 of these}$

The correct answer is C

When x   is    [ 3, 5 ]   then     l x - 4 l  ≤ 1   is true

When y is    [ 4, 8 ]  then   l x - 6 l  ≤  2    is true 

2^(x+y) * 3^(x-y) * 6^(2x+2y) = 72

$2^{x+y} * 3^{x-y} * 6^{2x+2y}= 72\\ 2^{x+y} * 3^{x-y} * 2^{2(x+y)}*3^{2(x+y)}= 3^2*2^3\\ 3^{x-y+2x+2y} * 2^{3(x+y)}= 3^2*2^3\\ 3^{3x+y} * 2^{3x+3y}= 3^2*2^3\\ so\\ 3x+y=2\quad(1a)\\ 3x=2-y \quad (1b)\\ 3x+3y=3\quad(2a)\\ x+y=1\quad(2b)\\ y=1-x\quad(2c)\\ sub (2c) into (1b)\\ 3x=2-(1-x)\\ 3x=1+x\\ 2x=1\\ {\bf{x=0.5,\quad y=0.5}} \\ check\\ 2^1*3^0*6^{2}=72\qquad \text{Das ist ausgezeichnet!}$

2^(x+y) * 3^(x-y) * 6^(2x+2y) = 72

2^(x + y) * 3^(x - y) * ( 6^2 )^(x + y)  =  72

2^(x + y) * 3^(x - y) * (36)^(x + y) = 72

[ 2 * 36]^(x + y) * 3^(x - y) = 72

[72]^(x + y) * 3^(x - y) =  72

If we let x, y =  1/2 we have that

2^1 * 3^0 * 6^2 = 72

2 * 1 * 36 = 72

So

(x, y) =  (1/2, 1/2)

6^(2x+2y) is 2^(2x+2y)*3^(2x+2y), so it is now 2^(3x+3y)*3^(3x+y)=72.

with some intuition, you can see that 2^3*3^2=72, so 3x+3y=3 and 3x+y=2.

now you subtract the second equation from the first so 2y=1 so y=1/2.

plugging back in, you can see that 1/2 is also x.

so the answer is (1/2, 1/2).

HOPE THIS HELPED!

For the first one, I define it as $-{1\over3} + {2 {5\over6}} - {3\over8} \cdot 1 {2\over3}.$ Now you calculate $-{1\over3}+{2 {5\over6}}$, which is $-{2\over6}+{2 {5\over6}}$. This is equal to ${2 {3\over6}}=2{1\over2}$.

Now we have to solve ${3\over8}\cdot1{2\over3}$. This is equal to ${3\over8}\cdot{5\over3}$, or $5\over8$. Now all you need to do is calculate $2{1\over2}-{5\over8}$, which is just $2{4\over8}-{5\over8}=1{12\over8}-{5\over8}=1{7\over8}.$

(The latex takes some time this took around 5 minutes lol)

For the second one, I define it as ${{-2{1\over2}}\over{-3{1\over3}}}+{({-{1\over6}})}$. Now we calculate the division, which is ${{5\over2}\over{10\over3}}.$ Since division is multiplying by a reciprocal, it is just ${5\over2}\cdot{3\over10}={3\over4}.$ Now you have to do ${3\over4}+(-{1\over6}),$ which becomes ${9\over12}+(-{2\over12})={7\over12}.$

For the third one, I define it as $-{1\over4}+(-{4\over5})\cdot{2{1\over6}}-{5\over16}.$ First we have to calculate the multiplication, which is ${-{4\over5}}\cdot{2{1\over6}}={-{4\over5}}\cdot{13\over6}={-{26\over15}}.$Now you do the addition and subtraction, which is ${-{1\over4}}-{26\over15}-{5\over16}={-{60\over240}}-{416\over240}-{75\over240}={-{551\over240}}.$

I really hope this helped, it took me around 20 minutes to solve this and do the latex.

HOPE THIS HELPED!!!!!!!!!!!!!!!!!!!!!!!!!!!

Can you help me with English tho?