Recent content by rtwikia

Never mind. Thanks anyway.:)

YES it worked!(Rock) $\log_{x}\left({10}\right)\cdot\log\left({x}\right)+(\log\left({x}\right))^2=2\cdot\log\left({x}\right)$ $\frac{1}{\log\left({x}\right)}\cdot\log\left({x}\right)+(\log\left({x}\right))^2=2\cdot\log\left({x}\right)$ $1+u^2=2u$ $u^2-2u+1=0$ $u=\frac{-(-2)\pm\sqrt{4-4}}{2}=1$...

Of course, if it is known that x=10 then I can do that. However I'm not sure that 10 is the answer, so...I'll just try to solve the quadratic(Nerd)

Should be 10?

The question is $$\log_{x}\left({10}\right)+log(x)=2$$ where (obviously) I have to find x. I tried changing the base, $\frac{log(10)}{log(x)}+log(x)=2$ $\frac{1}{log(x)}+log(x)=2$ ${log(x)}^{-1}+log(x)=log(100)$ but I could go no further. Whatever I try, I always got a wrong answer. By...

Okay then. And thanks by the way. I got the answer $0.27$ already.(Rock)

Yes! I got it!(Rock) $\beta=\frac{1}{2}\implies\alpha=2\implies c=2$ Thanks for your help!(Rock)

$\alpha$ and $\beta$ are the roots of the equation $2{x}^{2}-5x+c=0$. If $4\alpha-2\beta=7$, find the value of $c$. I did the following: $\alpha+\beta=-\frac{-5}{2}=\frac{5}{2}$ $\alpha\beta=\frac{c}{2}$ $\frac{c}{2}=\frac{7+2\beta}{4}\cdot\frac{-7+4\alpha}{2}$...

I've noted that it isn't given that $BFD$ is a straight line.:confused: Does it affect the deduction process?:confused:

After simplification, I found that $\overline{FG}=\frac{5x-1}{1+x}$. Can it be simplified further or is this the final answer? (When I measure the picture it measures nearly excatly $x$ cm):confused:

I got $\frac{1-x}{2}\left(1+\overline{FG}\right)=x\left(2-\overline{FG}\right)$ Am I right? :D

In the figure, ABCD is a square of side 1 cm. ABFE and CDFG are trapeziums(/trapeziods?). The Area of CDFG is twice the Area of ABFE. Let x cm be the length of AE. (a) Express the length of FG in terms of x. (b) Find the value of x, correct to 2 decimal places. Thanks! :D