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3 years ago

12

What are the next two terms in the sequence? 2, 14, 26, 38, ...

2 answers:

poizon [28]3 years ago

7 0

Answer:

<h2>50, 62</h2>

Step-by-step explanation:

$a(1)=2\\\\a(2)=a(1)+12=2+12=14\\\\a(3)=a(2)+12=14+12=26\\\\a(4)=a(3)+12=26+12=38\\\\a(5)=a(4)+12=38+12=50\\\\a(5)=a(5)+12=50+12=62$

If you want the formula.

It's an arithmetic sequence with the first term 2 and common difference 12.

The explicit formula of an arithmetic sequence:

$a(n)=a(1)+(n-1)d$

<em>d</em><em> - common difference</em>

Substitute:

$a(n)=2+(n-1)(12)$ <em>use the distributive property</em>

$a(n)=2+12n-12$

$a(n)=12n-10$

ki77a [65]3 years ago

5 0

Answer:

50, 62

The drawing will help

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2 years ago

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

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

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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,

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We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

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

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,

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

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Then,

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Hence largest volume,

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