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3 years ago

Write a rule for the nth term of the sequence 37, 32, 27, 22, . . . .

Mathematics

1 answer:

Radda [10]3 years ago

5 0

Answer:

a(n) = 37 - 5(n -1)

Step-by-step explanation:

Notice that each new term is obtained by subtracting 5 from the previous term. Thus, this is an arithmetic sequence with common difference -5 and first term 37.

The explicit formula for the nth term is a(n) = a(1) - 5(n - 1) = 37 - 5(n -1)

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A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e

ladessa [460]

Answer:

$\frac{r(1-\sqrt{2})}{-3}$

Step-by-step explanation:

Volume of cone = $\frac{1}{3} \pi r^{2} h$

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

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

= $\frac{1}{3} \pi r^{3}$

Surface area of cone including 1 base = $\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}$

Since r = h

So, area = $\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}$

= $\pi r^{2} +\pi\times r \times \sqrt{2r^2}$

= $\pi r^{2} +\pi\times r^2 \times \sqrt{2}$

Ratio of volume of cone to its surface area including base :

$\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}$

$\frac{\frac{1}{3}r}{1+\sqrt{2}}$

$\frac{r}{3(1+\sqrt{2})}$

Rationalizing

$\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\frac{r(1-\sqrt{2})}{-3}$

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is $\frac{r(1-\sqrt{2})}{-3}$

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Step-by-step explanation:

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Answer:

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2 years ago

What is the correct answer

krek1111 [17]

Answer:

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Step-by-step explanation:

3x - 9

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