Which property is illustrated by this problem 5(2 + 4) = (5.2) (5 . 4 )

Each day, enough paper is recycled in the United States to fill 15 miles of train boxcars. How many miles of boxcars could be fi

nikdorinn [45]

I think that it might be75

3 0

3 years ago

Read 2 more answers

The water level in the aquarium shark tank is always greater than 25 feet if the water level decreased by 6 feet during cleaning

stepan [7]

25 - 6 = 19 so 19 ft. it was 19 ft before he cleaned out the tank.

4 0

3 years ago

Cells are made up of?

Marianna [84]

Answer:

organaelles can i have brainlyest?

ik i didnt spell right

Step-by-step explanation:

6 0

3 years ago

Find any three rational numbers between 1/4 and 1/2

cluponka [151]

Answer:

12/40, 15/40, 18/40

Step-by-step explanation:

1/4, 1/2 = 1/4, 2/4

Since, Three rational numbers cannot come in between 1/4 and 2/4, we multiply them with 10.

∴ 1/4 × 10 = 10/40

∴ 2/4 × 10 = 20/40

∴ Three rational numbers between 10/40 and 20/40 = 12/40, 15/40, 18/40

(You can add more numbers from between 10/40 and 20/40 if you need)

7 0

3 years ago

Calculate the area of the region bounded by y=6x and y=6x^2. And find the volume of the solid generated by revolving the region

expeople1 [14]

Answer:

Hence, the Area is $\frac{24}{5} \;\text {and the Volume of the bounded region is } \frac{24\pi}{5}$

Step-by-step explanation:

Concept :

Area of the region bounded by two curve :

$\int_a^b(R^2-r^2)dx$

volume of the solid generated by revolving the region about the x-axis is :

$\int_a^b\pi(R^2-r^2)dx$ where $R$ is long radius of bounded region from x-axis and $r$ is short radius of bounded region from x-axis.

Given :

$y=6x$ and $y=6x^2$

To find :

Area of the region bounded by $y=6x$ and $y=6x^2$ and volume of the solid generated by revolving the region about the x-axis.

Explanation :

$\because y=6x\;\text{and}\;y=6x^2$

$\therefore 6x=6x^2\\\Rightarrow x=1$

Area of the region bounded by $y=6x$ and $y=6x^2$ is :

$\int_a^b(R^2-r^2)dx=\int_0^1((6x)^2-(6x^2)^2)dx\\$

$\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=\int_0^1(36x^2-36x^4)dx\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\int_0^1(x^2-x^4)dx\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left(\frac{1}{3}-\frac{1}{5}\right)\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\times\frac{2}{15}\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=\frac{24}{5}$

Volume of the solid generated by revolving the region about the x-axis.

$\int_0^1\pi((6x)^2-(6x^2)^2)dx=\int_0^1\pi(36x^2-36x^4)dx$

$\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\int_0^1(x^2-x^4)dx\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left(\frac{1}{3}-\frac{1}{5}\right)\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\times\frac{2}{15}\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=\frac{24\pi}{5}$

5 0

3 years ago