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3 years ago

Danisha found the volume of the figure shown.Her work is shown below.

Mathematics

2 answers:

stira [4]3 years ago

8 0

Answer:

C, just did the assignment.

Step-by-step explanation:

Mazyrski [523]3 years ago

4 0

Answer: C, She did not have the volume of the sphere to find the volume for the half sphere

Step-by-step explanation:

there wasn't a volume

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The volume of a cylinder is V = 1/3 πr2h. If the radius is doubled and the height is halved, what will be the ratio of the new v

ozzi

<span>V(o) = 1/3 πr²h - old volume
</span>V(n) = 1/3 π(2r)²(h/2)=1/3 π(2r)²(h/2)=1/3*4/2* πr²h=1/3*2* πr²h
V(n)/V(o)=1/3*2* πr²h/1/3 πr²h=2/1
V(n)/V(o)=2/1 - ratio
the volume will be doubled

8 0

3 years ago

Read 2 more answers

Plz answer both questions if works will give brainliest BTW start from the 1st one at the top

Nadya [2.5K]

Answer 1: 8

Answer 2: C

4 0

2 years ago

Read 2 more answers

And those... sorryyyyyy!:(

sasho [114]

Answer:

F: 18

Step-by-step explanation:

find total amount of points

43 + 7 = 50

subtract half of 50 to find the midpoint

43 - 25

L = 18

3 0

3 years ago

Read 2 more answers

25.7% as a fraction<br> plz help

vlada-n [284]

Answer:

257/1000

Step-by-step explanation:

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4 0

3 years ago

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|\\\\$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.

5 0

2 years ago