If f is the function given by f(x)= integral from (4 to 2x) of sqrt(t^2-t)dt then f'(2) =?

2 answers:

5 0

The question ask to calculate the function f'(2) in the F is the function given by f(x) = integral from (4 to 2x) of sqrt(t^2-t), base on that, the possible answer would be that the f(2) is 4 sqrt 3. I hope you are satisfied with my answer and feel free to ask for more if you have questions

sammy [17]3 years ago

3 0

Hello,

$f(x)= \int\limits^{2x}_4 {\sqrt{t^2-t} \, dt \\

 \dfrac{df(x)}{dx} = \dfrac{df(x)}{d(2x)} * \dfrac{d(2x)}{dx} =2* \sqrt{(2x)^2-2x} \\

f'(2)=2 * \sqrt{4^2-4} =2* \sqrt{12} =4 \sqrt{3}$

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A line passes through (-1, 7) and (2, 10) Which answer is the equation of the line?

Masteriza [31]

The answer to the question

4 0

3 years ago

In ΔLMN, \overline{LN}

Karolina [17]

Answer:

$\bold{\angle NLM = 10^\circ}$

Step-by-step explanation:

Given a ΔLMN.

Line LN is extended to point O.

such that:

$\text{m}\angle MNO = (5x-13)^{\circ}$

$\text{m}\angle NLM = (x-4)^{\circ}$ and

$\text{m}\angle LMN = (2x+19)^{\circ}$

To find:

$\text{m}\angle NLM=?$

Solution:

Kindly refer to the attached image for the given triangle and dimensions of angles.

Let us recall the external angle property of a triangle:

The external angle of a triangle is equal to the sum of two opposite internal angles.

i.e.

$\angle MNO = \angle NLM + \angle LMN\\\Rightarrow 5x-13 = x-4+2x+19\\\Rightarrow 5x-13 = 3x+15\\\Rightarrow 5x-3x = 13+15\\\Rightarrow 2x=28\\\Rightarrow x =14$