Which set contains only whole numbers divisible to 6?

1 answer:

7 0

Only <span>set Y = {6, 24, 36, 78, 84}, contains only whole numbers divisible by 6.

There were no options but this was the answer to this question on a quiz.

hope this helps!</span>

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The figure shows the design of a greenhouse with a rectangular floor. The front and back sides of the greenhouse are semicircles

Ksju [112]

9514 1404 393

Answer:

96 cubic feet

Step-by-step explanation:

A volume that is 16 ft by 18 ft by 1/3 ft will be ...

V = LWH

V = (16 ft)(18 ft)(1/3 ft) = 96 ft³

96 cubic feet of concrete are needed.

7 0

2 years ago

The football team has 2 large water containers. Each container holds 345π in3 of water. Each player is given a cylindrical water

raketka [301]

Answer:

They'll be able to get 34 bottles from the containers.

Step-by-step explanation:

Since the bottles are cylindrical we can calculate their volume by using the following formula:

V = base_area*h

V = \pi*(r^2)*h

r = d/2 = 4/2 = 2 inches

V = 3.14*(2^2)*5 = 3.14*4*5

V = 3.14*20 = 62.8 inches^3

In order to know how many full bottles the players will get we need to divide the total volume of the containers, which is given by the sum of the volume of each container, and divide it by the volume of each bottle. We have:

bottles = (345*pi + 345*pi)/62.8 = 690*pi/62.8 = 2,166.6/62.8 = 34.5

Since the problem wants the amount of full bottles we only take the integer part, so they will be able to get 34 bottles from the containers.

4 0

2 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$