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i think your just bad because this is coming from an 13 year old girl it is very easy if you put your mind to it

11 lolz

73/99  IS in simplified form       or in decimal form = .73737373....

One more chime in...... 

x = time of one pipe   x+2 = time of other pipe

Much like two resistors in parallel:   r= (r1xr2)/(r1+r2)

x * (x+2) / (x +(x+2))  = 6

(x^2 +2x)/ (2x+2) = 6

x^2 +2x = 12x + 12

x^2-10x-12 = 0           Now use quadratic formula

x = (10+- sqrt(100+48))/2          x = 5+- 6.08     x = 11.08 hrs      the other pipe is x+2 = 13.08 hrs

Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?

W ist die Entleerung des Tanks.

x ist die Zeit des Einen

x-2h ist die Zeit des Anderen

\(W=(\frac{W}{x}+\frac{W}{x-2h})\times 6h\)

\(\frac{1}{6h}= \frac{1}{x}+\frac{1}{x-2h}\)

\(\frac{1}{6h}=\frac{x-2h+x}{x^2-2hx}\)

\(\frac{x^2-2hx}{6h}=2x-2h\)

\(\Large To\ be \ continued\)

!

Not necessarily.  It depends if your calculator (or software) is set in degree or radian mode.

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Two pipes together drain a wastewater holding tank in 6 hours.
If used alone to empty the tank, one takes 2 hours longer than the other.
How long does each take to empty the tank when used alone?

\(\begin{array}{|rcll|} \hline \text{Let } q_1 &=& \text{Volumetric flow pipe } 1 \\ \text{Let } q_2 &=& \text{Volumetric flow pipe } 2 \\ \text{Let } V &=& \text{Volume of the tank} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & q_1 \cdot t + q_2 \cdot t &=& V \\ & (q_1 + q_2) \cdot t &=& V \quad & | \quad t = 6\ hours \\ (1) & (q_1 + q_2) \cdot 6 &=& V \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (2) & q_1\cdot t_1 &=& V \\ & t_1 &=& \frac{V}{q_1} \\ \hline & q_2 \cdot t_2 &=& V \quad & | \quad t_2 = t_1 + 2\ hours \\ (3) & q_2 \cdot (t_1 + 2) &=& V \\ & q_2 \cdot (t_1 + 2) &=& V \quad & | \quad t_1 = \frac{V}{q_1} \\ & q_2 \cdot (\frac{V}{q_1} + 2) &=& V \\ & \dots \\ (4) & V &=& \frac{2q_1q_2}{q_1-q_2} \\ \hline \end{array} \)

(1) = (4):

\(\begin{array}{|lrcll|} \hline & V = 6\cdot (q_1+q_2) &=& \frac{2q_1q_2}{q_1-q_2} \\ & 6\cdot (q_1+q_2) &=& \frac{2q_1q_2}{q_1-q_2} \quad & | \quad : 2\\ & 3\cdot (q_1+q_2) &=& \frac{q_1q_2}{q_1-q_2} \quad & | \quad \cdot (q_1-q_2) \\ & 3\cdot (q_1+q_2)(q_1-q_2) &=& q_1q_2 \\ & 3\cdot (q_1^2-q_2^2) &=& q_1q_2 \\ & 3q_1^2-3q_2^2-q_1q_2 &=& 0 \\ & 3q_1^2-q_1q_2-3q_2^2 &=& 0 \\\\ & q_1 &=& \frac{q_2\pm \sqrt{q_2^2-4\cdot 3 \cdot (-3q_2^2) } } {2\cdot 3} \\ & q_1 &=& \frac{q_2\pm \sqrt{ q_2^2+36q_2^2 } } {6} \\ & q_1 &=& \frac{q_2\pm \sqrt{ 37q_2^2 } } {6} \\ & q_1 &=& \frac{q_2\pm q_2\sqrt{37} } {6} \\ & q_1 &=& \frac16(1\pm\sqrt{37} ) q_2\\\\ & \mathbf{q_1 = \frac16(1+\sqrt{37} ) q_2 } &\text{ or }& q_1 = \frac16(1-\sqrt{37} )q_2 \\ & & & \text{ no solution } q_1 < 0 ! \\ (5) & \mathbf{\frac{q_1}{q_2} = \frac16(1+\sqrt{37} ) } \\ \hline \end{array}\)

(2) = (3):

\(\begin{array}{|lrcll|} \hline & V = q_1\cdot t_1 &=& q_2 \cdot (t_1 + 2) \\ & q_1\cdot t_1 &=& q_2 \cdot (t_1 + 2) \\ (6) & \frac{q_1}{q_2 } &=& \frac{t_1 + 2}{t_1} \\ \hline \end{array} \)

(5) = (6):

\(\begin{array}{|rcll|} \hline \frac{q_1}{q_2} = \frac16( 1+\sqrt{37} ) &=& \frac{t_1 + 2}{t_1} \\ \frac16( 1+\sqrt{37} ) &=& \frac{t_1 + 2}{t_1} \\ (1+\sqrt{37} )\cdot t_1 &=& 6\cdot(t_1 + 2) \\ (1+\sqrt{37} )\cdot t_1 &=& 6\cdot t_1 + 12 \\ (1+\sqrt{37} -6)\cdot t_1 &=& 12 \\ (\sqrt{37} - 5)\cdot t_1 &=& 12 \\ t_1 &=& \frac{12}{\sqrt{37} - 5} \\ t_1 &=& \frac{12}{6.08276253030- 5} \\ t_1 &=& \frac{12}{1.08276253030} \\ \mathbf{ t_1 } &\mathbf{=}& \mathbf{11.0827625303\ hours }\\\\ t_2 &=& t_1 + 2\ hours \\ \mathbf{ t_2 } &\mathbf{=}& \mathbf{13.0827625303\ hours }\\ \hline \end{array}\)

Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?

I always have an awful problem getting my head around questions like this so I made the question easier to see how it would work.

I said:

Let the tank be  36L     and let the times to drain be 4 hours and 6 hours respectively

So tank 1 will flow at 36/4 = 9L/hour

and tank 2 will flow at 36/6 = 6L/hour

if they are both draining then the water will flow at   (9+6) L/hour = 15L/hour

so 

\(\frac{15L}{1hour}=\frac{15L*\frac{36}{15}}{1hour*\frac{36}{15}}=\frac{36L}{2.4hours}\)

It will take 2.4 hours to drain the tank if both pipes are emptying it.

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Ok now I can do the same thing for your more difficult problem.

Let the tank be V litres

Pipe1 will take t hours to drain it so that is    \(\frac{V}{t}\;\;litres/hour\)

Pipe2 will take (t+2) hours to drain it so that is   \(\frac{V}{t+2}\;\;litres/hour\)

So the 2 pipes together will drain

       \(\frac{V}{t}+\frac{V}{t+2}\;\;litres/hour\\ =V[\frac{1}{t}+\frac{1}{t+2}]\;\;litres/hour\\ =V[\frac{(t+2)+t}{t(t+2)}]\;\;litres/hour\\ =\frac{V\left[\frac{2t+2}{t^2+2t}\right]\;\;litres}{1hour}\\ =\frac{V\left[\frac{2t+2}{t^2+2t}\right]\left[\frac{t^2+2t}{2t+2}\right]\;\;litres}{\left[\frac{t^2+2t}{2t+2}\right]hour}\\ =\frac{V\;\;litres}{\left[\frac{t^2+2t}{2t+2}\right]hour}\\\)

So it will take    

\(\frac{t^2+2t}{2t+2}\;\;\text{hours to drain the tank}\\~\\ now\\ \frac{t^2+2t}{2t+2}=6\\ t^2+2t=6(2t+2)\\ t^2+2t=12t+12\\ t^2-10t-12=0\\ t=\frac{10\pm\sqrt{100+48}}{2}\\ t=\frac{10\pm\sqrt{148}}{2}\\ t=\frac{10\pm2\sqrt{37}}{2}\\ t=5\pm\sqrt{37}\\ \text{t cannot be negative so }\\t=5+\sqrt{37}\approx 11.08276253\approx 11hours\;\;4\;minutes \;and\;58\;seconds\\\)

Which is near enough to 11 hours and 5 minutes

So the pipes will take 11 hours and 5 minutes and 13 hours and 5 min respectively to empty the tank on their own.

Since they take 6 hours together that sounds reasonable    

810

I would be interested in this too

Use prime factorization to find the cube root of 3375

\(3375=3^3\times 125=3^3\times5^3=(3\times5)^3=15^3\)

\(\sqrt[3]{3375}=15\)

!

Use prime factorization to find the cube root of 3375

\(\begin{array}{|rcll|} \hline && \sqrt[3]{3375} \\ &=& \sqrt[3]{3^3\cdot 5^3} \\ &=& \sqrt[3]{3^3} \cdot \sqrt[3]{ 5^3} \\ &=& 3 \cdot 5 \\ &=& 15 \\ \hline \end{array}\)

3*3*3*5*5*5

Find the sum of 35 terms of an arithmetic series
of which the first term is a and the fifteenth term is 9a.

Let t₁ = a

Let t₁₅ = 9a

Formula:

\(\begin{array}{|lrcll|} \hline (1) & t_x &=& t_1 + (x-1)\cdot d \\ (2) & t_y &=& t_1 + (y-1)\cdot d \\ (3) & t_z &=& t_1 + (z-1)\cdot d \\ \hline I = (1) - (2) & t_x-t_y &=& d\cdot (x-y) \\ II = (1) - (3) & t_x-t_z &=& d\cdot (x-z) \\ \hline II / I & \frac{t_x-t_z}{t_x-t_y} &=& \frac{d\cdot (x-z)}{d\cdot (x-y)} \\ & \frac{t_x-t_z}{t_x-t_y} &=& \frac{x-z}{x-y} \quad & | \quad \cdot (t_x-t_y) \\ & t_x-t_z &=& (t_x-t_y)\cdot \frac{x-z}{x-y} \\ & t_z &=& t_x-(t_x-t_y)\cdot \frac{x-z}{x-y} \\ & t_z &=& t_x- t_x\cdot (\frac{x-z}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) \\ & t_z &=& t_x\cdot (1-\frac{x-z}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) \\ & t_z &=& t_x\cdot (\frac{x-y-x+z}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) \\ &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ \hline \end{array}\)

t₃₅ = ?

\(\begin{array}{|lrclrcl|} \hline x = 1 \quad & t_x &=& t_1 = a \\ y = 15 \quad & t_y &=& t_{15} = 9a \\ z = 35 \quad & t_z &=& t_{35} = \ ? \\ &&&& \mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &&&& t_{35} & = & a\cdot ( \frac{35-15}{1-15}) + 9a\cdot (\frac{1-35}{1-15}) \\ &&&& t_{35} & = & a\cdot (\frac{20}{-14}) + 9a\cdot (\frac{-34}{-14}) \\ &&&& t_{35} & = & -a\cdot (\frac{10}{7}) + 9a\cdot (\frac{17}{7}) \\ &&&& t_{35} & = & 9a\cdot (\frac{17}{7}) -a\cdot (\frac{10}{7}) \\ &&&& t_{35} & = & a\cdot (\frac{9\cdot 17}{7}) -a\cdot (\frac{10}{7}) \\ &&&& t_{35} & = & a\cdot (\frac{153}{7}) -a\cdot (\frac{10}{7}) \\ &&&& t_{35} & = & a\cdot (\frac{153-10}{7}) \\ &&&& \mathbf{ t_{35} } &\mathbf{=}& \mathbf{ \frac{143}{7}a }\\ \hline \end{array} \)

s₃₅ = ?

\(\begin{array}{|rcll|} \hline s_{35} &=& \left(\frac{t_1+t_{35}}{2} \right)\cdot 35 \quad & | \quad t_1 = a \quad t_{35} = \frac{143}{7}a \\ s_{35} &=& \left(\frac{a+\frac{143}{7}a}{2} \right)\cdot 35 \\ s_{35} &=& \left(\frac{1+\frac{143}{7} }{2} \right)\cdot 35 a\\ s_{35} &=& \left( \frac{150}{14} \right)\cdot 35 a\\ s_{35} &=& 150\cdot \frac{5}{2} a\\ s_{35} &=& 75\cdot 5 a\\ \mathbf{ s_{35} } &\mathbf{=}& \mathbf{375 a}\\ \hline \end{array} \)

In mathematics MAD is Mean Absolute Deviation.  It is less sensitive to outliers than the standard deviation.

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y=1634, you get it by dividing 55556 by 34

We need math so that we can count the amount of dank memes on the web.

what is 1/2+1/4+1/8

\(\begin{array}{|rcll|} \hline && \frac12 + \frac14 + \frac18 \\ &=& \frac12 + \frac{1}{2^2} + \frac{1}{2^3} \\ &=& \frac12 \cdot ( 1 + \frac{1}{2} + \frac{1}{2^2} ) \\ &=& \frac12 \cdot [ 1 + \frac{1}{2} \cdot ( 1+ \frac{1}{2} ) ] \\ &=& \frac12 \cdot ( 1 + \frac{1}{2} \cdot \frac{3}{2} ) \\ &=& \frac12 \cdot ( 1 + \frac{3}{4} ) \\ &=& \frac12 \cdot \frac{7}{4} \\ &=& \frac{7}{8} \\ \hline \end{array}\)

-138 mod 116

\(\begin{array}{|rcll|} \hline -138 \equiv x \pmod {116} \\ -138+116 \equiv x \pmod {116} \quad & | \quad 116 \pmod {116} = 0 \\ -22+116 \equiv x \pmod {116} \quad & | \quad 116 \pmod {116} = 0 \\ 94 \equiv x \pmod {116} \\\\ -138 \pmod {116} = 94 \pmod {116} = 94 \\ \hline \end{array} \)

maybe 755/355

625 plus 130 = 755

225 plus 130 = 335

what is the surface area of a cylinder with the radius of 5 and height of 11

Let A = surface area
Let r = radius
Let h = height

\(\begin{array}{|rcll|} \hline A &=& (2\pi r)\cdot h + 2\cdot (\pi r^2) \\ A &=& (2\pi r)\cdot (h + r) \quad & | \quad r = 5 \qquad h = 11 \\ A &=& (2\pi 5)\cdot (11 + 5) \\ A &=& 10\cdot \pi\cdot 16 \\ A &=& 160\cdot \pi \\ A &=& 502.654824574 \\ \hline \end{array}\)

Sorry.....wrong units in final answer ...should be  ft-lb    instead of  j  

Was dinkin' around with this.....and thought maybe since we are pumping from the top (6ft) to the bottom....we could just use the average of 3 ft to calculate work:

Total volume of water =   pi x r^2 x 6 = 301.592 ft^3  

Total weight of water = 301.592 x 62.4 = 18.819.396 lb

Total work to move the water 3 ft  = 18,819.396 x 3 = 56458.189 ft-lb     Which basically agrees with the previous answer

3600 + 9*2150 + 17*1300 + 16*775 + 18*460 + 68*275 + 65*165 + 19*100 = 97055

Please help me find the next three numbers in this geometric sequence

625,225,75...  

and please explain how you got that answer so i can understand.

A geometric  sequence has the same factor  between each number. Not here !

\(\frac{625}{225} = 2.77777777778 \\ \frac{225}{75} = 3 \\ 2.77777777778 \ne 3 \)