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3 years ago

Jason had 3/4 pound of salami. He used 1/3 pound for sandwiches.

1 answer:

4 0

<span>Jason had ¾ pound of Salami.
He used 1/3 of it for sandwhich, now find the number of pound of salami left.
=> ¾ - 1/3
=> ¾ - 1/3, since the given equation contains different denomination, find the least common denominator
=> so the LCD is 12
=> ¾ - 1/3 = 9/12 – 4/12
=> 9/12 – 4/12
=> 5/12
Thus, the answer is letter C. 5/12 pounds is left to Jason’s Salami.</span>

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The first thing I would probably do is distribute (multiply) the -5/2 into the parenthesis:

= (-5/2)x + (-20/2)

Notice how I am not putting the x into the fraction. This is to make it easier to see the slope-intercept form.

Lets simplify the (-20/2) and then add 7 to both sides:

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Your answer would then be:

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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:

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

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

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

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

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