All categories

3 years ago

How to differentiate cbrt(8+9xsqrt4)

Mathematics

1 answer:

shusha [124]3 years ago

3 0

Answer:

gay

Step-by-step explanation:

gay gay gay gay gay math sucks do drugs

You might be interested in

A triangle has an area of 48 square units. Its height is 8 units.

wariber [46]

12 units
(because 0,5 x 12 x 8 = 48)

5 0

3 years ago

Read 2 more answers

The diameter of a planet is about 1420 mi. Find the volume of the planet. Round to the nearest thousand cubic miles. Use 3.14 fo

Ainat [17]

Answer:

1,498,454,000 sq. miles

Step-by-step explanation:

Volume of a sphere = 4/3 (3.14) r^3

So since the diameter is 1420 miles, the radius would be half.

V = 4/3 x (3.14) x 710^3

V=4.18666667 x 357,911,000

V = 1,498,454,054.5 but if you want it rounded to the nearest thousand then it'd be 1,498,454,000.

8 0

3 years ago

Mr. Katz's science class is planning a field trip. Thirty people are going on the trip. There are 7 drivers and two types of veh

Fudgin [204]

6 vans will be taken to the trip

3 0

3 years ago

(14+13n+3n^3)+(2+4n+9n^3)

Sliva [168]

Answer:

=12n3+17n+16

i hope this helps :)

8 0

3 years ago

Read 2 more answers

\int (x+1)\sqrt(2x-1)dx

Nezavi [6.7K]

Answer:

$\int (x+ 1) \sqrt{2x-1} dx = \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C$

Step-by-step explanation:

$\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v = \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}$

$\int (x+1)\sqrt(2x-1)dx\\\\ = uv - \int v du$

$= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ u = x + 1 => du = dx \ ]$

$= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\$

6 0

3 years ago