2 different sets of equations:
F+2T=30
(0.5F^ -0.5)/(0.5T^ -0.5)=1/2
4F+2T=30
(0.5F^ -0.5)/(0.5T^ -0.5)=2
$\displaystyle F+2T=30$
$\displaystyle
\frac{{(0.5F)}^ \frac{-1}{2}}{{(0.5T)}^ \frac{-1}{2}}=\frac{1}{2}
$
If the above is your question
Cancel 0.5 of numerator and denominator and your question becomes
$\displaystyle
\sqrt{\frac{T}{F}} =\frac{1}{2}
$
Squaring both sides
$\displaystyle
T=\frac{F}{4}
$
So
$\displaystyle
T =\frac{30-2T}{4}
$
I think you can go ahead and solve this and next question
Hello schoolhelpFirst of all, notice that in $\displaystyle \frac{0.5F^{-0.5}}{0.5T^{-0.5}}$ you can cancel the initial 0.5's to get $\displaystyle \frac{F^{-0.5}}{T^{-0.5}}$
Then you need to know that:
- $\displaystyle a^{-x} = \frac{1}{a^x}$; and
- $\displaystyle a^{0.5}=\sqrt{a}$
So $\displaystyle F^{-0.5} = \frac{1}{F^{0.5}}$
$\displaystyle = \frac{1}{\sqrt{F}}$
Do the same with the denominator:
$\displaystyle T^{-0.5}=\frac{1}{\sqrt{T}}$
and now invert and multiply, and the second equation reads
$\displaystyle \frac{\sqrt{T}}{\sqrt{F}}=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{T}{F} = \frac{1}{4}$
$\displaystyle \Rightarrow F=4T$
Substitute into the first equation and solve for $\displaystyle T$, and then $\displaystyle F$.
Do the second problem in a similar way.
OK now?
Grandad