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KonstantinChe [14]

3 years ago

11

What's is the converse of the statement?If x is odd, then 2x is even.

1 answer:

dsp733 years ago

7 0

If you have two statements p and q and $p\Rightarrow q$ is true, then $\neg q\Rightarrow \neg p$ is also true.

In your case, statements are p - "x is odd" and q - "2x is even".

Then $\neg p$ - "x is not odd" and $\neg q$ - "2x is not even."

Hence $\neg q\Rightarrow \neg p$ sounds as: "If 2x is not even, then x is not odd".
Answer: correct choice is D.

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The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

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$V=\frac{1}{3}\pi r^2 h$

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$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

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$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

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