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What's wrong with you

1. is 0.725

Thanks so much everyone!

Perimeter =2[L + W]

Area =L x W

2[L + W] = 40
L x W =96, solve for L, W

L=12  and  W = 8 - feet, or:

W = 12  and  L = 8 - feet.

Very good!!.

wow..thanks Heureka

Help Me Please!!

System of Equation: solve the system and express answers as ordered pairs

{x²+y²=34}

{x+y=8}

\(\begin{array}{|rclrcl|} \hline x^2+y^2 &=& 34 & x + y &=& 8 \\ && & y &=& 8 -x \\\\ x^2 +(8-x)^2 &=& 34 \\ x^2 + 64-16x+x^2 &=& 34 \\ 2x^2 -16x + 64-34 &=& 0 \\ 2x^2 -16x + 30 &=& 0 \quad & | \quad :2 \\ x^2 -8x + 15 &=& 0 \\\\ x_{1,2} &=& \dfrac{8 \pm \sqrt{64-4\cdot 15} }{2} \\\\ &=& \dfrac{8 \pm \sqrt{4} }{2} \\\\ &=& \dfrac{8 \pm 2 }{2} \\\\ &=& 4\pm 1 \\\\ x_1 &=& 4+1 = 5 \\ x_2 &=& 4-1 = 3 \\\\ y_1 &=& 8-x_1 =8-5 = 3 \\ y_2 &=& 8-x_2= 8-3 = 5 \\ \hline \end{array}\)

\(x_1 = 5 \qquad x_2 = 3 \\ y_1 = 3 \qquad y_2 = 5 \)

Thanks Heureka,

I found my own error too :)

It was just a careless sign error in my last line.

\(x=log_2(({\frac{1-y}{y}})^{-1})\\ x=-log_2\left({\frac{y}{1-y}}\right) \qquad \text{This should have been } \color{red}{x=+log_2\left({\frac{y}{1-y}}\right) }\\ \therefore f^{-1}(x)=\color{red}{+} log_2\left({\frac{x}{1-x}}\right)\\ \text{Using the change of base rule this is exactly the same as Heureka's answer.} \)

Your answer from the internet is no doubt correct, it also matches Heureka's answer I think.

BUT

Write x^2 + 6x - 7 in the form of (x + a)^2 + b
where a and b are integers

\(\begin{array}{|rcll|} \hline && (x+a)^2+b \\\\ &=& x^2+\underbrace{2a}_{=6}x + \underbrace{a^2 +b}_{=-7} \\\\ 2a = 6 && a^2+b =-7 \\ a = 3 && 3^2+b = -7 \\ && b =-7-9 \\ && b = -16 \\\\ && \mathbf{(x+3)^2-16} \\ \hline \end{array}\)

Find the inverse of the function

\(\displaystyle f(x)=\frac{2^x}{1+2^x}\)

f(x)=\frac{2^x}{1+2^x}

\(\begin{array}{|rcll|} \hline y &=& \dfrac{2^x}{1+2^x} \quad & | \quad \text{substitute $u =2^x$ } \\\\ y &=& \dfrac{u}{1+u} \\\\ y(1+u) &=& u \\ y+yu &=& u \\ y &=& u-yu \\ y &=& u(1-y) \\\\ u&=& \dfrac{y}{1-y} \quad & | \quad \text{substitute $u =2^x$ } \\\\ 2^x&=& \dfrac{y}{1-y} \quad & | \quad \text{ $\ln{}$ both sides } \\\\ \ln(2^x) &=& \ln\left( \dfrac{y}{1-y} \right) \\\\ x\ln(2) &=& \ln\left( \dfrac{y}{1-y} \right) \\\\ x &=& \dfrac{ \ln\left( \dfrac{y}{1-y} \right) } {\ln(2) } \quad & | \quad x \Leftrightarrow y \\\\ \mathbf{f^{-1}} &\mathbf{=}&\mathbf{ \dfrac{ \ln\left( \dfrac{x}{1-x} \right) } {\ln(2) } } \\ \hline \end{array}\)