All Answers 357330 Answers

Just plug in the values.

x is 2 ... so 4x is (4 times 2) ... = 8.     Hold that thought.

y is 3 ... so y² is (3 times 3) ... = 9, ... then 1/3 of that = 3.

Add them together (8 plus 3) is 11.  And that's your answer. 

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Now that is funny yeah I belive it is a Female the name 

(Melody)

rogelio necesita un tramo de tubo con una longitud que esté exactamente a la mitad de un tubo de 1 1/2m y otro de 2 5/8 ¿cuál es la longuitud del tubo?

Find the largest value of
$\dfrac{y}{x}$
for pairs of real numbers
$(x,y)$
that satisfy
$(x - 3)^2 + (y - 3)^2 = 6$.

$\text{Let $OC^2 = x_c^2+y_c^2 = 3^2+3^2 = 2\cdot 3^2 $}$

$\begin{array}{|rcll|} \hline OT^2 +r^2 &=& OC^2 \\ OT^2 + 6 &=& 2\cdot 3^2 \\ OT^2 &=& 18 - 6 \\ &=& 12 \\ &=& 3\cdot 4 \\ \mathbf{OT} & \mathbf{=} & \mathbf{2\sqrt{3}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \tan(C) &=& \dfrac{y_c}{x_c} \\ &=& \dfrac{3}{3} \\ \mathbf{\tan(C) } &\mathbf{=}&\mathbf{1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \tan(B) &=& \dfrac{r}{OT} \\ &=& \dfrac{\sqrt{6}}{2\sqrt{3}} \\ &=& \dfrac{\sqrt{2}\sqrt{3}}{2\sqrt{3}} \\ \mathbf{\tan(B) } &\mathbf{=}&\mathbf{\dfrac{\sqrt{2} }{2 }} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \tan(A) &=& \tan(B+C) \\\\ &=& \dfrac{\tan(B)+\tan(C)} {1-\tan(B)\tan(C)} \\\\ &=& \dfrac{1+\dfrac{\sqrt{2} }{2 }} {1-\dfrac{\sqrt{2} }{2 }} \\\\ &=& \dfrac{2+ \sqrt{2} } {2- \sqrt{2} } \\\\ &=& \dfrac{2+ \sqrt{2} } {2- \sqrt{2} }\cdot \left( \dfrac{2+ \sqrt{2} } {2+ \sqrt{2} } \right) \\\\ &=& \dfrac{\left(2+ \sqrt{2}\right)^2 } {4-2 } \\\\ &=& \dfrac{6+4\sqrt{2}} {2 } \\\\ \mathbf{\tan(A) } &\mathbf{=}&\mathbf{3+2\sqrt{2} } \\ \hline \end{array}$

The largest value of $\dfrac{y}{x} = \tan(A) = 3+2\sqrt{2}$

$\text{This one I'd do using Lagrange multipliers}\\ f(a,b,c,d) = a \cdot \sqrt{b} \cdot \sqrt[3]{c} \cdot \sqrt[4]{d}\\ g(a,b,c,d) = 36a+4b+4c+3d-25\\ \text{Solve}\\ \nabla(f-\lambda g) = 0$

$y = \pm \sqrt{6-(x-3)^2}+ 3\\ \dfrac{y}{x} = \dfrac{ \pm \sqrt{6-(x-3)^2}+ 3}{x}\\ \text{The maximum will be when the numerator has a positive sign}\\ \dfrac{y}{x} = \dfrac{ \sqrt{6-(x-3)^2}+ 3}{x}\\ \dfrac{d}{dx} \dfrac{y}{x} = -\dfrac{3 (x-1)}{x^2 \sqrt{-x^2+6 x-3}}\\ \text{and this clearly equals 0 at }x=1$

$\text{The second derivative evaluated at 1 is less than zero so this is a maximum}$

$\dfrac{y}{x} = \sqrt{6-(1-3)^2}+3 = 3+\sqrt{2}$

Thanks!

$\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}\\ n! = n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 3 \cdot 2$

$9x^2 +54x - (y^2 - 10y)+55 = \\ 9(x^2 + 6x +9 - 9)-(y^2-10y +25-25) + 55=\\ 9(x+3)^2 - 81 - (y-5)^2 +25 + 55 = \\ 9(x+3)^2 - (y-5)^2 = 1\\ \dfrac{(x+3)^2}{\left(\frac 1 3\right)^2} - (y-5)^2 = 1$

$\text{The vertices are at }\\ (x,y)=(-3\pm \dfrac 1 3 ,5)=\left (-\dfrac{10}{3},5\right),~\left(-\dfrac{8}{3},5\right)\\ \text{and thus they are }\dfrac 2 3 \text{ units apart}$

The digits aren't ADDED....they are used to form  an eight digit number

First digit   1- 6

Second digit  1-6

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eighth digit 1-6       no digits of the eight will be > 6

Thanks guys!!!

But aren't you counting the case where the last three digits are 128?

As a male I don't think I'd quickly forgive my parents for naming me Melody.

$4x+2 \text{ is the sum of }4x \text{ and }2\\ 7y \text{ is a term, (there are others)}\\ 5(4x+2) \text{ is the product of }5 \text{ and }(4x+2)\\ \text{There is no factor present}\\ \text{There is no quotient present}\\ 7 \text{ is a coefficient of y}$

$\text{on the line }y=x, \text{ oddly enough for every point }y=x\\ \text{thus we have}\\ \dfrac{x^2-x}{2x-x+2} = a\\ x^2-x=a x + 2a\\ x^2 -(1+a)x -2a=0\\ x = \dfrac{(1+a)\pm \sqrt{(1+a)^2+8a}}{2}$

$\text{You then have two points of the form }(x_k,x_k),~k=1,2\\ \text{Where the }x_k \text{ are the roots above}\\ \text{Find the distance between these two points. It's }\\ d=\sqrt{2} \sqrt{\left | a^2+10 a+1\right| }\\ \text{Solve for }d=6 \text{ and choose the value of }a \text{ that is positive}\\ \text{and leads to real points }x,y\\ a=\sqrt{42}-5 \text{ is the only one}$

1. Degree 11 + Degree 7   will result in a polynomial of Degree 11

2.  We only  need to multiply the  terms of each polynomial that have the highest degree...so.....

x^4 *  2x^4   =   2x^8    .....so....degree 8

Look at x = 2     draw a vertical line at   x = 2   this crosses the graph at    y = 2     

Look for the point where   y = 4   crosses the graph ....this occurs at x = 3

Maybe we need a "net"   ????

The BLUE graph is the same as the RED graph BUT it is shifted two units to the LEFT

y = f(x+2)