Hello forum, Vairon here. I'm having trouble rewriting any kind of formula but I can hopefully understand the first one and use it as reference. Solve for r $\displaystyle C&=2&\pi&r$ I assume C/2r*pi should solve it
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Not exactly. If you divide both sides by $\displaystyle 2 \pi r$, you get "1" on the right hand side, not r! So just divide by $\displaystyle 2 \pi $ to get $\displaystyle r = \frac{C}{2 \pi}$
Originally Posted by TheChaz Not exactly. If you divide both sides by $\displaystyle 2 \pi r$, you get "1" on the right hand side, not r! So just divide by $\displaystyle 2 \pi $ to get $\displaystyle r = \frac{C}{2 \pi}$ I feel proud I actually got it before the response thanks though btw just to know I'm doing everything right I got this one answered. Area of a triagle Solve for b : $\displaystyle A&=\frac{1}{2}&bh$ I solved it : $\displaystyle \frac{A}{\frac{1}{2}*h}&=b$
$\displaystyle b = \frac{2A}{h}$
Originally Posted by skeeter $\displaystyle b = \frac{2A}{h}$ Will mine work? I tried plugin in the numbers and both formulas output the same number
Originally Posted by vaironxxrd Will mine work? I tried plugin in the numbers and both formulas output the same number it'll work, but you're expected to be able to clear any fraction "inside" another fraction.
Originally Posted by skeeter it'll work, but you're expected to be able to clear any fraction "inside" another fraction. I'm sorry to ask this but why is it 2? Flip and multiply?
Originally Posted by vaironxxrd I'm sorry to ask this but why is it 2? Flip and multiply? Roughly speaking, yes.
Originally Posted by vaironxxrd I feel proud I actually got it before the response thanks though btw just to know I'm doing everything right I got this one answered. Area of a triagle Solve for b : $\displaystyle A&=\frac{1}{2}&bh$ I solved it : $\displaystyle \frac{A}{\frac{1}{2}*h}&=b$ $\displaystyle \displaystyle \begin{align*} A &= \frac{1}{2}bh \\ 2A &= 2\cdot \frac{1}{2}bh \\ 2A &= bh \\ \frac{2A}{h} &= \frac{bh}{h} \\ \frac{2A}{h} &= b \end{align*}$
Originally Posted by vaironxxrd Solve for b : $\displaystyle A&=\frac{1}{2}&bh$ or start off by multiplying both sides by 2: 2A = bh ; then: b = 2A / h Done.
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