All Answers 357280 Answers

Two ends are not needed. The water isn't going to get out from the top.

Two ends are not needed. The water isn't going to get out from the top.

The graph below should help:

4 Dollars spent. Tacos are 70 cents each, burritos are 40 cents each

4 Dollars spent. Tacos are 70 cents each, burritos are 40 cents each

Hint:

You can write this as $1000^3-\sum_{k=1}^{999}k^3$

Now $\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$

Hope this helps!

Sheet metal n the tank:

TWO ends have area of pi r^2     2  x pi 4^2

PLUS the circumference x height =  pi x d x 30 = pi x 8 x 30      Add reds together to get surface area of sheet metal.

Volume of cylinder = pi r^2 h   = pi  4^2 x 30 = ......

Utah question does not have map diagram.....

i meant that 123456789 all are +and- can 0 be one two

∑[(n^4+3n^2+10n+10) / (2^n(n^4+4)), n, 2,∞] = 11 / 10

um no I what to no if 0 can be -.

HiylinLink

Do  you mean possible? If so -0 is not possible. 0 means there is no amount of something there. For this answer it would be a negative number. 

yes and no will 1+1-1*1/1=1 always yes or no is so then explained and why does it go back to one. 

Multiply both sides by g, divide both sides by sin(84°), take the square root of the result.

How to rearrange the formula to get the u?

"find log25(8) * log1/2(5)"

.

"When a long jumper leaves the ground at an an angle of 42.0 above the horizontal. He covers a distance of 9.00m. What is the take off speed of the jumper?"

Let

$n$

and

$k$

be positive integers such that $\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11$. Find the smallest possible value of .

$\begin{array}{|rcll|} \hline \dfrac{\dbinom{n}{k}} {\dbinom{n}{k + 1}} &=& \dfrac{4}{11} \quad | \quad \dbinom{n}{k + 1} = \dbinom{n}{k}\cdot \dfrac{n-k}{k+1} \\\\ \dfrac{\dbinom{n}{k}} {\dbinom{n}{k}\cdot \dfrac{n-k}{k+1}} &=& \dfrac{4}{11} \\\\ \dfrac{k+1}{n-k} &=& \dfrac{4}{11} \\\\ 11(k+1) &=& 4(n-k) \\ 11k+11 &=& 4n-4k \\ \mathbf{4n-15k} & \mathbf{=} & \mathbf{11} \\\\ 4n &=& 11+15k \\ n &=& \dfrac{11+15k} {4} \\ &=& \dfrac{11+1-1+15k+k-k} {4} \\ &=& \dfrac{12+16k-(1+k)} {4} \\ &=& \dfrac{12+16k} {4}- \dfrac{(1+k)} {4} \\ &=& 3+4k - \underbrace{\dfrac{(1+k)} {4}}_{=a} \\ \mathbf{n} &\mathbf{=}& \mathbf{3+4k - a} \\\\ a &=& \dfrac{1+k} {4} \\ 4a &=& 1+k \\ \mathbf{k} &\mathbf{=} & \mathbf{4a-1,\quad a\in \mathbb{Z}}\\\\ n &=& 3+4k - a \quad | \quad k=4a-1 \\ &=& 3+4(4a-1) - a \\ &=& 3+16a-4 - a \\ \mathbf{n} &\mathbf{=}& \mathbf{-1+15a} \\\\ n_{min} &=& -1+15\cdot 1\quad | \quad a=1 \\ \mathbf{n_{min}} &\mathbf{=}& \mathbf{14} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline k&=& 4a-1 \quad | \quad a=1 \\ k&=& 4\cdot1 -1 \\ k&=& 3 \\ \dfrac{\dbinom{14}{3}} {\dbinom{14}{4}} &=& \dfrac{4}{11} \\ \hline \end{array}$

2. 

∫5 3 (1/(x^2-6x+13))dx

$\large{\int \limits_{3}^{5} \dfrac{1}{ \sqrt{x^2-6x+13} }\ dx}$

$\begin{array}{|rcll|} \hline && \mathbf{ \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{x^2-6x+13} }\ dx} \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{(x-3)^2-9+13} }\ dx \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{(x-3)^2+4} }\ dx \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1} { \sqrt{ 4 \left( 1+ \dfrac{ (x-3)^2 } {4} \right) } }\ dx \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1} { 2 \sqrt{ 1+ \left(\dfrac{x-3}{2}\right)^2 } }\ dx \\\\ &=& \dfrac{1}{2} \displaystyle \int \limits_{3}^{5} \dfrac{1} {\sqrt{ 1+ \left(\dfrac{x-3}{2}\right)^2 } }\ dx \\\\ && \quad \boxed{ \text{substitute: } \\ \dfrac{x-3}{2} = \sinh(\theta),\ \dfrac{1}{2}dx=\cosh(\theta)d\theta,\ dx=2\cosh(\theta)d\theta \\ x=3 \Rightarrow \theta=\sinh^{-1}\left(\dfrac{3-3}{2}\right)=0,\\ x=5\Rightarrow \theta=\sinh^{-1}\left(\dfrac{5-3}{2}\right)\\ =\sinh^{-1}(1)=\ln(1+\sqrt{1^2+1})=0.88137358702} \\\\ &=& \dfrac{1}{2} \displaystyle \int \limits_{0}^{0.88137358702} \dfrac{1} {\sqrt{ 1+ \Big(\sinh(\theta)\Big)^2 } }2\cosh(\theta)\ d\theta \\\\ &=& \displaystyle \int \limits_{0}^{0.88137358702} \dfrac{\cosh(\theta)} { \cosh(\theta) }\ d\theta \\\\ &=& \displaystyle \int \limits_{0}^{0.88137358702} d\theta \\\\ &=& \Big[ \theta \Big]_{0}^{0.88137358702} \\\\ &=& 0.88137358702 \\\\ && \mathbf{ \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{x^2-6x+13} }\ dx = \sinh^{-1}(1) = 0.88137358702} \\ \hline \end{array}$

1. 

∫1/6 0 (1/(sqrt(1-9x^2))dx

$\large{\int \limits_{0}^{\frac{1}{6}} \dfrac{1}{ \sqrt{1-9x^2} }dx}$

$\begin{array}{|rcll|} \hline && \mathbf{ \large{\int \limits_{0}^{\frac{1}{6}} \dfrac{1}{ \sqrt{1-9x^2} }\ dx} } \\\\ &=& \displaystyle \int \limits_{0}^{\dfrac{1}{6}} \dfrac{1}{ \sqrt{1-(3x)^2} }\ dx \\\\ && \quad \boxed{ \text{substitute: } \\ u = 3x,\ du=3\ dx,\\ x=0 \Rightarrow u=3\cdot 0=0,\\ x=\dfrac{1}{6}\Rightarrow u=3\cdot \dfrac{1}{6}=\dfrac{1}{2} } \\\\ &=& \displaystyle \int \limits_{0}^{\dfrac{1}{2}} \dfrac{1}{ \sqrt{1-u^2} } \dfrac{du}{3} \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{1}{2}} \dfrac{1}{ \sqrt{1-u^2} }\ du \\\\ && \quad \boxed{ \text{substitute: } \\ u = \sin(\theta),\ du=\cos(\theta)d\theta,\\ u=0 \Rightarrow \theta=\arcsin(0)=0,\\ u=\dfrac{1}{2}\Rightarrow \theta=\arcsin\left(\dfrac{1}{2}\right)=\dfrac{\pi}{6}} \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{\pi}{6}} \dfrac{\cos(\theta)}{ \sqrt{1-\Big(\sin(\theta)\Big)^2} }\ d\theta \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{\pi}{6}} \dfrac{\cos(\theta)}{ \cos(\theta) }\ d\theta \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{\pi}{6}} d\theta \\\\ &=& \dfrac{1}{3} \Big[ \theta \Big]_{0}^{\frac{\pi}{6}} \\\\ &=& \dfrac{1}{3}\cdot \dfrac{\pi}{6} \\\\ &\mathbf{=}& \mathbf{\dfrac{\pi}{18}} \\ \hline \end{array}$

I was trying to solve the following question:

$\begin{array}{|rcll|} \hline x^2(x+3y)-y^3 &=& 3 \\ x^3 + 3x^2y-y^3 - 3 &=& 0 \\\\ \mathbf{f(x,y)} & \mathbf{=} & \mathbf{x^3 + 3x^2y-y^3 - 3} \\\\ \dfrac{\partial f} {\partial x} &=& 3x^2+6xy \\ \dfrac{\partial f} {\partial y} &=& 3x^2-3y^2 \\ && \boxed{\text{Formula: }\\ \dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f} {\partial x}} {\dfrac{\partial f} {\partial y}} } \\ \dfrac{dy}{dx} &=& -\dfrac{ 3x^2+6xy } { 3x^2-3y^2 } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ 3(x^2+2xy) } { 3(x^2-y^2) } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ x^2+2xy } { x^2-y^2 } \\\\ \mathbf{ \dfrac{dy}{dx}} &\mathbf{=}& \mathbf{\dfrac{ x^2+2xy } { y^2-x^2 }} \\ \hline \end{array}$