Which of the following methods of timekeeping is the least precise?

Find the volume of the cylinder with a diameter of 12 inches and a height of 10

kati45 [8]

Answer:

360 pi in ^3

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h

We know the diameter is 12 so the radius is 1/2 the diameter

r = d/2 = 12/2 = 6

V = pi (6)^2 * 10

V = pi (36)*10

V = 360 pi in ^3

We can approximate pi by 3.14

V =1130.4 in ^3

Or we can approximate pi by using the pi button

V =1130.973355 in ^3

6 0

3 years ago

----------- are used to represent an unknown quantity in a mathematical expression.

anastassius [24]

A letter or symbol is used to represent an unknown quantity

4 0

3 years ago

Read 2 more answers

What is There are fourteen whole snack bars on a plate. There are five friends.

AleksandrR [38]

Carl is incorrect. Dave ate a higher fraction of snack bars, by 0.2 snack bars.

Carl had .5 left of a snack bar.

Dave had .3 left of a snack bar.

Tony had .5 left of a snack bar.

Gary had .0 left of a snack bar.

Tryone had .7 left of a snack bar.

If we add the above snack bars, there is a total of two remaining snack bars, meaning they only ate 12 of 14 snack bars.

3 0

3 years ago

the cost of renting a van is $30. An additional $0.89 is charged for each mile, m, the van is driven, Which expression can be us

kolbaska11 [484]

Answer:

30 + 0.89m

Step-by-step explanation:

I think this is the correct answer.

5 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago