All categories

3 years ago

A cylindrical soup can has a radius of 1.2 in. and is 6 in. tall. What's the volume?

Mathematics

1 answer:

Sergio [31]3 years ago

7 0

Answer:

Volume= 27. 12inches³

Step-by-step explanation:

to solve this question lets write out the parameters given

radius of the cylinder = 1.2 inches

height of the cylinder = 6 inches

volume of the cylinder = ?

<u>solution</u>

the fomular to find the volume of a cylinder is given to be

Volume= πr²h

Volume= 3.14 × ( 1.2)² × 6

Volume= 3.14 × 1.44 × 6

Volume= 27. 12inches³

therefore the volume of the cylinder equals to 27. 12inches³

You might be interested in

Find the volume of the prism. 4 in. 8 in. 8 in.

Temka [501]

Answer:

256

Step-by-step explanation:

8×8=64

64×4=256

256

3 0

2 years ago

Read 2 more answers

How do I find the reciprocal of the fractions 2/3 and 8/3

Doss [256]

To find the reciprocal of a number, just flip the numerator and the denominator around.

A. The reciprocal of 2/3 is 3/2.

C. Firstly, convert the mixed fraction into an improper fraction. Multiply the whole number by the denominator and add the numerator to that product. After that, put the quantity on the numerator. In this case, 6 x 6 = 36 and 36 + 1 = 37, so our improper fraction is 37/6. Next, just flip the numerator and denominator around, and your reciprocal is 6/37.

4 0

3 years ago

Adam wrote a check for $38 to pay his monthly gas bill, but when balancing his checkbook, he accidently recorded it as a credit

Alex73 [517]

It would be $76 over, I hope this helps.

7 0

4 years ago

Read 2 more answers

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$