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7 wrz 2017 · If we want to solve V = 1 3 πr2h for h, we need to isolate the term with h (already done), and then multiply both sides by the inverses of everything other than h. The multiplication by 3 πr2 to both sides is our choice; we do this so that each piece other than h has a multiplicative inverse that cancels it off.
prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx \sum_{n=0}^{\infty}\frac{3}{2^n} Show More
Divide each term in r2h = 3V π r 2 h = 3 V π by r2 r 2 and simplify. Tap for more steps... Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.
Połączenie (ang. self join), w którym relacja łączy się "sama z sobą" (występuje kilkukrotnie w klazuli FROM). Użycie aliasów w połączeniu zwrotnym jest obowiązkowe.
By similar triangles, observe that: \dfrac{h}{3}=\dfrac{r}{2} \iff r=\dfrac{2h}{3} Hence, substituting into the formula for the volume of a cone will help us to avoid product rule: V=\dfrac{1}{3}\pi \left(\dfrac{2h}{3}\right)^2h = \dfrac{4\pi}{27}h^3 ...
By similar triangles, observe that: \dfrac{h}{3}=\dfrac{r}{2} \iff r=\dfrac{2h}{3} Hence, substituting into the formula for the volume of a cone will help us to avoid product rule: V=\dfrac{1}{3}\pi \left(\dfrac{2h}{3}\right)^2h = \dfrac{4\pi}{27}h^3 ...
For this one, V = (1/3)(pi)(r^2)(h) we divide both sides by (1/3)(pi)(r^2) to get: V/[(1/3)(pi)(r^2)]=h Got it? RJ wwwmath-unlock.com
