All categories

3 years ago

The formula for the volume of a cylinder is V = pir2h. Solve the formula for the height h.

Mathematics

1 answer:

RideAnS [48]3 years ago

5 0

H=v/(2pir)

You just have to isolate h. To do that you are going to divide both sides by 2pir.

You might be interested in

A costume designer bought 1.4 kg of sequince. She used 250 grams to make a crown and 400 grams to make a cape. How many grams of

Salsk061 [2.6K]

Find the total amount in grams.

1 kg = 1000 g

1.4 * 1000 = 1400

So she bought a total of 1400 grams.

Add up the grams she used:

400 + 250 = 650

Subtract it from the total:

1400 - 650 = 750

So 750 grams of sequence were left.

5 0

3 years ago

Read 2 more answers

25 miles per hour = how many feet per second

ollegr [7]

Answer:36.67 i think

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the fraction of 4.8

kenny6666 [7]

To get the improper fraction of 4.8 you can first get the mixed number of 4 and 4/5. You must then convert 4 into fifths and add the two fractions together 4 times 5 is twenty so you must add 20/5 plus 4/5 and the improper fraction you end up with is 24/5.

4 0

3 years ago

Read 2 more answers

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago

Rob laid three sticks end to end. The lengths of the sticks were 45 millimeters, 32 centimeters, and 4 decimeters. How long, in

Oksanka [162]

Answer:

0.765 meters long were the sticks when laid end to end.

Step-by-step explanation:

Given:

Rob laid three sticks end to end. The lengths of the sticks were 45 millimeters, 32 centimeters, and 4 decimeters.

Now, to find the length in meters of the sticks when laid end to end.

The lengths of the sticks are:

1st stick - 45 millimeters.

2nd stick - 32 centimeters.

3rd stick - 4 decimeters.

So, we convert the units of sticks into meters by using conversion factor:

1st stick:

1 millimeter = 0.001 meter.

45 millimeters = 0.001 × 45

45 millimeters = 0.045 meter.

2nd stick:

1 centimeter = 0.01 meter.

32 centimeters = 0.01 × 32

32 centimeters = 0.32 meter.

3rd stick:

1 decimeter = 0.1 meter.

4 decimeters = 0.1 × 4

4 decimeters = 0.4 meter.

Now, adding all the three sticks length to get the the length in meters of the sticks when laid end to end:

$0.045+0.32+0.4\\\\=0.765\ meters.$

Therefore, 0.765 meters long were the sticks when laid end to end.

5 0

4 years ago