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lilavasa [31]
3 years ago
8

A cone is placed inside a cylinder. The cone has half the radius of the cylinder, but the height of each figure is the same. The

cone is tilted at an angle so its peak touches the edge of the cylinder’s base. What is the volume of the space remaining in the cylinder after the cone is placed inside it?
Mathematics
2 answers:
kirza4 [7]3 years ago
7 0
Volume of cylinder = pi x r^2 x h
<span>Volume of cone = 1/3 x pi x r^2 x h
</span>radius of cone = 1/2 (r)
Remaining volume = Volume of cylinder - <span>Volume of cone 
                               = </span>pi x r^2 x h - <span>1/3 x pi x (r/2)^2 x h
                               = </span><span>pi r^2  h - 1/4 pi r^2 h
                               = 3/4 (</span><span>pi r^2  h</span>)
Dmitriy789 [7]3 years ago
7 0

3/4 (pi r^2 h) is the correct answer


I hope this helps!!

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Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0
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Answer:

\dfrac{-1}{6}

Step-by-step explanation:

Given the limit of a function expressed as \lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)

Step 2: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the function

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\

Step 3: substitute x = 0 into the resulting function

= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)

Step 4: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\

=  \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)

Step 6: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\

Step 7: substitute x = 0 into the resulting function in step 6

=  \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}

<em>Hence the limit of the function </em>\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \  is \ \dfrac{-1}{6}.

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