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2 years ago

Integral of 1/(x^2sqrt(4x+1))?

Mathematics

1 answer:

cestrela7 [59]2 years ago

3 0

<span>Integral of 1/(x^2sqrt(4x+1)) = 4tanh^-1(sqrt(4x + 1)) - sqrt(4x + 1) / x + c</span>

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nlexa [21]

Given:

Equation of line $7x-4y=-12$ .

To find:

The equation of line that goes through the point ( − 21 , 2 ) and is perpendicular to the given line.

Solution:

The given equation of line can be written as

$7x-4y+12=0$

Slope of line is

$\text{Slope}=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}$

$m_1=-\dfrac{7}{(-4)}$

$m_1=\dfrac{7}{4}$

Product of slopes of two perpendicular lines is -1. So, slope of perpendicular line is

$m_1m_2=-1$

$m_2=-\dfrac{1}{m_1}$

$m_2=-\dfrac{4}{7}$ $[\because m_1=\dfrac{7}{4}]$

Now, the slope of perpendicular line is $m_2=\dfrac{4}{7}$ and it goes through (-21,2). So, the equation of line is

$y-y_1=m_2(x-x_1)$

$y-2=-\dfrac{4}{7}(x-(-21))$

$y-2=-\dfrac{4}{7}x-\dfrac{4}{7}(21)$

$y-2=-\dfrac{4}{7}x-12$

$y=-\dfrac{4}{7}x-12+2$

$y=-\dfrac{4}{7}x-10$

Therefore, the required equation in slope intercept form is $y=-\dfrac{4}{7}x-10$ .

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3 years ago

Find the area of shaded part .

Firlakuza [10]

Answer:

Area of big rectangle= 5m× 3m

= 15m²

Area of small rectangle=1m× 1m

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Area of shaded part= Area of big rectangle- Area of a small rectangle

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Step-by-step explanation: Hope it helps

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3 years ago

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saul85 [17]

Answer:

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