3 quater of number is 15 work out four fifths of the number

Mathematics

1 answer:

just olya [345]3 years ago

4 0

Answer:

Step-by-step explanation:

15×3=45÷5=8×4=32.

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There are 64 cookies in a jar. The probability of randomly choosing oatmeal cookies from the jar is 37.5%. How many of the cooki

Rufina [12.5K]

Answer:

the number of cookies that are oatmeal cookies is 24

Step-by-step explanation:

Given that

There are 64 cookies in a jar

And, the probability of randomly selecting the oatmeal cookies is 37.5%

We need to find out the number of cookies that are oatmeal cookies

So, the number of cookies that are oatmeal cookies are

= 64 cookies × 37.50%

= 24 cookies

hence, the number of cookies that are oatmeal cookies is 24

6 0

2 years ago

A. n=55<br>b. n=25<br>c. n=5<br>d. n=3025

Mnenie [13.5K]

Answer:

Step-by-step explanation:

Assuming the triangle is right, use Pythagoras' identity to find N

The square on the hypotenuse is equat to the sum of the squares on the other 2 sides.

Here the hypotenuse = 73, thus

N² + 48² = 73², that is

N² + 2304 = 5329 ( subtract 2304 from both sides )

N² = 3025 ( take the square root of both sides )

N = $\sqrt{3025}$ = 55 → A

5 0

3 years ago

\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:

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

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

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

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