Ask question

Ask question

All categories

3 years ago

6

Find the volume and surface area of a sphere with radius 1 yards. Round your answer to two decimal places.

1 answer:

olga55 [171]3 years ago

8 0

Answer:

Volume of the sphere up to two decimal places= $4.19$ cubic yards

Surface area of the sphere up to two decimal places $=12.56$ square yards

Step-by-step explanation:

Given:

The radius of the sphere= 1 yard

Volume of a Sphere:

$\frac{4}{3}* \pi *r^3$ : $\pi =\frac{22}{7} =3.14$

$=\frac{4}{3}*3.14*1*1*1$

$=4.187$ cubic yards

Volume of the sphere up to two decimal places= $4.19$ cubic yards

Surface area of the sphere= $4*\pi* r^2$

$=4*3.14*1*1$

Surface area of the sphere up to two decimal places $=12.56$ square yards

You might be interested in

Use the Distributive Property to expand the expression -(u-v)

deff fn [24]

Answer:

(U+V) is the answer to the solution

7 0

3 years ago

lawyer [7]

Answer - A

Explanation - 8 - 3 = 5 50 - 25 = 25

8 0

3 years ago

Read 2 more answers

Which point lies on the y-axis?

NeTakaya

Answer:

Step-by-step explanation:

3 0

3 years ago

Amelia drives 10 miles in 20 minutes. If she drove three hours in total at the same rate, how far did she go? subquestion - how

olga nikolaevna [1]

Answer:

Amelia drove 90 miles in 3 hours.

Step-by-step explanation:

she drove 10 miles in 20 minutes

we multiply this by 3, and get

30 miles in an hour ( 60 minutes)

then we multiply this by 3 and get

90 miles in 4 hours (180 minutes) :)

5 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago