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I think it is meant.

Note, Max, that by inspection, at least one real solution  is {x,y}  = {1, 1}

10/30 + 12/30 + 15/30 = 37/30 = 1 7/30

If a conical tank with a radius of 8 feet, and a height of 12 feet is being filled with water a rate of 2 cubic feet of water per minute, find how fast the water level is rising when the depth of the water in the tank is 2 feet. 

Notice that the ratio of the height to the radius = 12/8  = 3/2

So....at any time, t,    r = (2/3)h

And we can  represent the volume as :

V =  (1/3)pi * r^2  * h  =   (1/3) [ (2/3)h] ^2 *h =  (1/3)pi (4/9)h^3  =  (4/27)pi*h^3

Take the derivative with respect to time, t

dv/dt  = 3*(4/27)pi*h^2* dh/dt

dv/dt = (12/27)pi* h^2 * dh/dt

dv/dt  = (4/9)pi*h^2* dh/dt

And dv/dt = 2ft^3/min.....and we need to solve for dh/dt  when h = 2ft

So we have

2ft^3/min  = (4/9)pi (2ft)^2 *  dh/dt

2ft^3/min  = pi*(16/9)ft^2 * dh/dt

[2] / [pi*(16/9)] ft/min = dh/dt  = 9 / [8 pi] ft/min  ≈ 0.358 ft/min

If a conical tank with a radius of 8 feet, and a height of 12 feet is being filled with water a rate of 2 cubic feet of water per minute, find how fast the water level is rising when the depth of the water in the tank is 2 feet.

Simplify the following:
(15 x^2-(6 x^2)/3-3 x-9 x-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

The gcd of 6 and 3 is 3, so 6/3 = (3×2)/(3×1) = 3/3×2 = 2:
(15 x^2-2 x^2-3 x-9 x-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

Grouping like terms, 15 x^2-2 x^2-3 x-9 x-6 = (15 x^2-2 x^2)+(-9 x-3 x)-6:
(15 x^2-2 x^2)+(-9 x-3 x)-6 (6 x^2-(12 x^2)/2+2 x-6 x-4)

15 x^2-2 x^2 = 13 x^2:
(13 x^2+(-9 x-3 x)-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

-9 x-3 x = -12 x:
(13 x^2+-12 x-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

The gcd of 12 and 2 is 2, so 12/2 = (2×6)/(2×1) = 2/2×6 = 6:
(13 x^2-12 x-6) (6 x^2-6 x^2+2 x-6 x-4)

Grouping like terms, 6 x^2-6 x^2+2 x-6 x-4 = (-6 x+2 x)-4+(6 x^2-6 x^2):
(-6 x+2 x)-4+(6 x^2-6 x^2) (13 x^2-12 x-6)

2 x-6 x = -4 x:
(13 x^2-12 x-6) (-4 x-4+(6 x^2-6 x^2))

6 x^2-6 x^2 = 0:
-4 x-4 (13 x^2-12 x-6)

Factor -4 out of -4 x-4:
Answer: |-4 (x+1) (13x^2 - 12x - 6)

How dare you! :(

Ask this calculator!! 

\((15x^2-9x-\dfrac{6}{3x^2}-3x-6)(6x^2-6x-\dfrac{12}{2x^2}+2x-4)\\ =(-(4x)-4)(13x^2-(12x)-6)\)

Hope that is correct......

Just when you are working in this my friend sent me help too:

His answer( He is good at maths too :) )

\(\text{Let }a = \sqrt x\\ \sqrt{2+3\sqrt x - x}+\sqrt{6 - 2\sqrt x - 3x}=\sqrt{ 10 + 4\sqrt x - 5x}\\ \sqrt{-a^2+3a+2}+\sqrt{-3a^2-2a+6}=\sqrt{-5a^2+4a+10}\\ \sqrt{(2-a^2)+3a}+\sqrt{3(2-a^2)-2a}=\sqrt{5(2-a^2)+4a}\\ \text{Let } b=2-a^2\\ \sqrt{b+3a}+\sqrt{3b-2a}=\sqrt{5b+4a}\\ \text{Let }u=\sqrt{b+3a}\text{, }v=\sqrt{3b-2a}\\ \therefore b+3a=u^2\text{, }3b-2a=v^2\\ a=\dfrac{3u^2-v^2}{11},b =\dfrac{2u^2+3v^2}{11} \\ \therefore 5b+4a=2u^2+v^2\\ \text{Substitute back into the equation}\\ u+v=\sqrt{2u^2+v^2}\\ u^2+2uv+v^2=2u^2+v^2\\ u^2-2uv=0\\ u(u-2v)=0\\ \text{If u = 0, b= -3a and b = }2-a^2\\ a=1\text{ OR }a=2(rejected)\\ \text{If u - 2v = 0}\\ u^2=4v^2\\ \therefore a=b\text{ OR }b = 2-a^2\\ \therefore a= 1\text{ Or }a=-2(Rejected)\\ \therefore x= 1\)

Don't even know what is he trying to do....... Is he wrong? Because he got only 1 of your answers!!

Solve for x:
2 sqrt((6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)) = 2+3 sqrt(x)-x

Raise both sides to the power of two:
4 (6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x) = (2+3 sqrt(x)-x)^2

Subtract (2+3 sqrt(x)-x)^2 from both sides:
4 (6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)-(2+3 sqrt(x)-x)^2 = 0

4 (6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)-(2+3 sqrt(x)-x)^2 = 44+44 sqrt(x)-77 x-22 x^(3/2)+11 x^2:
44+44 sqrt(x)-77 x-22 x^(3/2)+11 x^2 = 0

Simplify and substitute y = sqrt(x):
 44+44 sqrt(x)-77 x-22 x^(3/2)+11 x^2 = 44+44 sqrt(x)-77 sqrt(x)^2-22 sqrt(x)^3+11 sqrt(x)^4 = 11 y^4-22 y^3-77 y^2+44 y+44 = 0:
11 y^4-22 y^3-77 y^2+44 y+44 = 0

The left hand side factors into a product with four terms:
11 (y-1) (y+2) (y^2-3 y-2) = 0

Divide both sides by 11:
(y-1) (y+2) (y^2-3 y-2) = 0

Split into three equations:
y-1 = 0 or y+2 = 0 or y^2-3 y-2 = 0

Add 1 to both sides:
y = 1 or y+2 = 0 or y^2-3 y-2 = 0

Substitute back for y = sqrt(x):
sqrt(x) = 1 or y+2 = 0 or y^2-3 y-2 = 0

Raise both sides to the power of two:
x = 1 or y+2 = 0 or y^2-3 y-2 = 0

Subtract 2 from both sides:
x = 1 or y = -2 or y^2-3 y-2 = 0

Substitute back for y = sqrt(x):
x = 1 or sqrt(x) = -2 or y^2-3 y-2 = 0

Raise both sides to the power of two:
x = 1 or x = 4 or y^2-3 y-2 = 0

Add 2 to both sides:
x = 1 or x = 4 or y^2-3 y = 2

Add 9/4 to both sides:
x = 1 or x = 4 or y^2-3 y+9/4 = 17/4

Write the left hand side as a square:
x = 1 or x = 4 or (y-3/2)^2 = 17/4

Take the square root of both sides:
x = 1 or x = 4 or y-3/2 = sqrt(17)/2 or y-3/2 = -sqrt(17)/2

Add 3/2 to both sides:
x = 1 or x = 4 or y = 3/2+sqrt(17)/2 or y-3/2 = -sqrt(17)/2

Substitute back for y = sqrt(x):
x = 1 or x = 4 or sqrt(x) = 3/2+sqrt(17)/2 or y-3/2 = -sqrt(17)/2

Raise both sides to the power of two:
x = 1 or x = 4 or x = (3/2+sqrt(17)/2)^2 or y-3/2 = -sqrt(17)/2

Add 3/2 to both sides:
x = 1 or x = 4 or x = (3/2+sqrt(17)/2)^2 or y = 3/2-sqrt(17)/2

Substitute back for y = sqrt(x):
x = 1 or x = 4 or x = (3/2+sqrt(17)/2)^2 or sqrt(x) = 3/2-sqrt(17)/2

Raise both sides to the power of two:
x = 1 or x = 4 or x = (3/2+sqrt(17)/2)^2 or x = (3/2-sqrt(17)/2)^2

2 sqrt((6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)) ⇒ 2 sqrt((6-2 sqrt(1)-3 1) (2+3 sqrt(1)-1)) = 4
2+3 sqrt(x)-x ⇒ 2+3 sqrt(1)-1 = 4:
So this solution is correct

2 sqrt((6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)) ⇒ 2 sqrt((6-2 sqrt(4)-3 4) (2+3 sqrt(4)-4)) = 4 i sqrt(10) ≈ 12.6491 i
2+3 sqrt(x)-x ⇒ 2+3 sqrt(4)-4 = 4:
So this solution is incorrect

2 sqrt((6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)) ⇒ 2 sqrt((6-2 sqrt((3/2-sqrt(17)/2)^2)-3 (3/2-sqrt(17)/2)^2) (2+3 sqrt((3/2-sqrt(17)/2)^2)-(3/2-sqrt(17)/2)^2)) = 2 sqrt(273-63 sqrt(17)) ≈ 7.27856
2+3 sqrt(x)-x ⇒ 2+3 sqrt((3/2-sqrt(17)/2)^2)-(3/2-sqrt(17)/2)^2 = 3 (sqrt(17)-3) ≈ 3.36932:
So this solution is incorrect

2 sqrt((6-2 sqrt(x)-3 x) (2+3 sqrt(x)-x)) ⇒ 2 sqrt((6-2 sqrt((3/2+sqrt(17)/2)^2)-3 (3/2+sqrt(17)/2)^2) (2+3 sqrt((3/2+sqrt(17)/2)^2)-(3/2+sqrt(17)/2)^2)) = 0
2+3 sqrt(x)-x ⇒ 2+3 sqrt((3/2+sqrt(17)/2)^2)-(3/2+sqrt(17)/2)^2 = 0:
So this solution is correct

The solutions are:
Answer: |x = 1                   or                       x = (3/2+sqrt(17)/2)^2

Use decimals

Just enter the fractions........rhe calculator can handle all four basic operations

Thanks CPhill :D

I am doing great! Thank you :)

It's not that easy.......

And your first step is already wrong.

\(\left (\sqrt{2+3\sqrt x - x}\;+\;\sqrt{6-2\sqrt x - 3x}\right )^2\) does not equal \(8+\sqrt x - 4x\)

√ [ 2 + 3√x - x]  + √ [6 - 2√x - 3x ]   = √10 + 4√x - 5x]

Note, Max......that, by inspection, x = 1   is at least one solution.......let's work through this

Square both sides

[ 2 + 3√x - x]  + 2√ [ 2 + 3√x - x] * √ [6 - 2√x - 3x ] + [6 - 2√x - 3x ]   = 10 + 4√x - 5x

Simplify

8 + √x - 4x  + 2 √ [( 2 + 3√x - x) (6 - 2√x - 3x) ] =   10 + 4√x - 5x

2 √ [( 2 + 3√x - x) (6 - 2√x - 3x) ]  =  2 + 3√x -x

2 √  [-7 x^(3/2)+3 x^2-18 x+14 sqrt(x)+12 ]   = 2 + 3√x -x

Square both sides again

4 [-7 x^(3/2)+3 x^2-18 x+14 sqrt(x)+12 ]  = -6 x^(3/2)+x^2+5 x+12 sqrt(x)+4

-28x^(3/2) + 12x^2 - 72x + 56sqrt(x) + 48  =  -6 x^(3/2)+x^2+5 x+12 sqrt(x)+4

22x^(3/2) - 11x^2 + 77x - 44sqrt(x) - 44  =  0

Divide through by 11

2x^(3/2) - x^2 + 7x - 4sqrt(x) - 4  = 0

This is a little difficult to factor......WolframAlpha gives

-(sqrt(x)-1) (sqrt(x)+2) (x-3 sqrt(x)-2) = 0

Setting the first factor to 0  →  sqrt(x) - 1  = 0       and it's obvious that  x = 1

Setting the second factor to 0  → sqrt (x) + 2  = 0  →  sqrt (x)  = -2    and this is a non-real result

Setting the third factor to 0  →  x - 3sqrt(x) - 2  = 0

And WolframAlpha  gives  [ 13 + 3√17 ] / 2    as the other real solution

So....the real solutions are  x = 1   and x = [ 13 + 3√17 ] / 2

Solve for x:
8+sqrt(x)-4 x = 10+4 sqrt(x)-5 x

Subtract 10+4 sqrt(x)-5 x from both sides:
-2-3 sqrt(x)+x = 0

Simplify and substitute y = -3 sqrt(x):
 -2-3 sqrt(x)+x = -2--3 sqrt(x)+((-3 sqrt(x))^2)/(9) = y^2/9-y-2 = 0:
y^2/9-y-2 = 0

Multiply both sides by 9:
y^2-9 y-18 = 0

Add 18 to both sides:
y^2-9 y = 18

Add 81/4 to both sides:
y^2-9 y+81/4 = 153/4

Write the left hand side as a square:
(y-9/2)^2 = 153/4

Take the square root of both sides:
y-9/2 = (3 sqrt(17))/2 or y-9/2 = -(3 sqrt(17))/2

Add 9/2 to both sides:
y = 9/2+(3 sqrt(17))/2 or y-9/2 = -(3 sqrt(17))/2

Substitute back for y = -3 sqrt(x):
-3 sqrt(x) = 9/2+(3 sqrt(17))/2 or y-9/2 = -(3 sqrt(17))/2

Divide both sides by -3:
sqrt(x) = -3/2-sqrt(17)/2 or y-9/2 = -(3 sqrt(17))/2

Raise both sides to the power of two:
x = (-3/2-sqrt(17)/2)^2 or y-9/2 = -(3 sqrt(17))/2

Add 9/2 to both sides:
x = (-3/2-sqrt(17)/2)^2 or y = 9/2-(3 sqrt(17))/2

Substitute back for y = -3 sqrt(x):
x = (-3/2-sqrt(17)/2)^2 or -3 sqrt(x) = 9/2-(3 sqrt(17))/2

Divide both sides by -3:
x = (-3/2-sqrt(17)/2)^2 or sqrt(x) = sqrt(17)/2-3/2

Raise both sides to the power of two:
x = (-3/2-sqrt(17)/2)^2 or x = (sqrt(17)/2-3/2)^2

8+sqrt(x)-4 x ⇒ 8+sqrt(((-3)/2-(sqrt(17))/(2))^2)-4 (-3/2-sqrt(17)/2)^2 = -11/2 (3+sqrt(17)) ≈ -39.1771
10+4 sqrt(x)-5 x ⇒ 10+4 sqrt(((-3)/2-(sqrt(17))/(2))^2)-5 (-3/2-sqrt(17)/2)^2 = -11/2 (3+sqrt(17)) ≈ -39.1771:
So this solution is correct

8+sqrt(x)-4 x ⇒ 8+sqrt(((sqrt(17))/(2)-(3)/2)^2)-4 (sqrt(17)/2-3/2)^2 = 13/2 (sqrt(17)-3) ≈ 7.30019
10+4 sqrt(x)-5 x ⇒ 10+4 sqrt(((sqrt(17))/(2)-(3)/2)^2)-5 (sqrt(17)/2-3/2)^2 = 19/2 (sqrt(17)-3) ≈ 10.6695:
So this solution is incorrect

The solution is:
Answer: |x = (-3/2-sqrt(17)/2)^2

\(\dfrac{(2x^3)(x^4)^2}{8x^{11}}\\ =\dfrac{2x^3(x^8)}{8x^{11}}\\ =\dfrac{2x^{11}}{8x^{11}}\\ =\dfrac{1}{4x^{0}}\\ =\dfrac{1}{4}\)

When you duel with a problem with variables but not an equation and the answer is a number, you can conclude that the original problem must be equal that answer.

i.e. \(\dfrac{(2x^3)(x^4)^2}{8x^{11}}\equiv \dfrac{1}{4}\)

Then I ask you, what is \(\dfrac{2\times(123123123)^3\times \left((123123123)^4\right)^2}{8\times(123123123)^{11}}\)You can tell that is 1/4!! Yeah this is great :)

\(-7\sqrt{y^2}-8\\ \text{Radicand = }y^2\\ \text{Index = }2\)

If you want to simplify:

\(-7\sqrt{y^2}-8\\ =-7y - 8\)

Example: The circumference(perimeter of circle) of a circle is 22 centimeters. Find the radius.

Let r cm be the radius.

\(2\times \pi \times r = 22\\ \pi \times r = 11\\ r=\dfrac{11}{\pi}\)

The radius is 11/pi cm.

I really need the steps........

\(\dfrac{1}{3}+\dfrac{1}{5}\\ =\dfrac{5}{15}+\dfrac{3}{15}\\ =\dfrac{8}{15}\)

I have a quicker way to do fraction addition actually, you can try to learn that or you can just ignore that if you want.

Diagonally multiply and add them up, put it at the numerator's place, multiply the denominators, simplify.

Don't understand? Never mind, Look at the example:

Example:

\(\dfrac{2}{3}+\dfrac{4}{5}\)

Diagonally multiply and add them up: 2 x 5 + 3 x 4 = 10 + 12 = 22

Put it at the numerator's place: we have \(\dfrac{22}{something\; here}\) now.

Multiply the denominators: 3 x 5 = 15. We now have \(\dfrac{22}{15}\), it is improper!!

Therefore we need to Simplify: Final answer = 1 7/15 :D You can check the answer by using the old way :)

x=1, or

x = 13/2+(3 sqrt(17))/2 (assuming a complex-valued square root)

From: W/A!!.

\(\dfrac{21}{57}\)

This fraction is not beautiful, why? Because it is not the simplest form :(

Then we will try to make both numerator and denominator smaller and at the same time keep its value the same.

But how??

We will divide it by a fraction which denominator and numerator is the same because if the denominator and numerator is the same, the fraction is equal to 1!

\(\dfrac{21}{57}\div \dfrac{3}{3}\)

Then we will distribute the division sign......

\(\dfrac{21\div 3}{57\div 3}\)

Then divide........

\(\dfrac{7}{19}\)

Yay that fraction is beautiful now because that we cannot do the same process to the fraction. It is because 7 and 19 have no common factors except 1.

What if we have this?

\(\dfrac{42}{126}\)

Repeat the progress with 2 over 2 as they are both divisible by 2!

\(\dfrac{42}{126}\div \dfrac{2}{2}\)

Distribute the division sign:

\(\dfrac{42\div 2}{126\div 2}\)

Divide them:

\(\dfrac{21}{63}\)

This is still not beautiful enough! We have to make it more beautiful!! :(

We can see that they are both divisible by 7, therefore we will repeat the process using 7 over 7.

\(\dfrac{21}{63}\div\dfrac{7}{7}\)

Distribute the division sign:

\(\dfrac{21\div 7}{63\div 7}\)

Divide them:

\(\dfrac{3}{9}\)

NO!! That's not beautiful enough :( 

Repeat the process with 3 over 3.

\(\dfrac{3}{9}\div \dfrac{3}{3}\)

Distribute the division sign:
\(\dfrac{3\div 3}{9\div 3}\)

Divide them:

\(\dfrac{1}{3}\)

Yay that's beautified by me again :D 1 and 3 have no common factors except 1!!