3 years ago

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Mathematics

1 answer:

Len [333]3 years ago

8 0

I just took the test!

Answer:

16.2 m²

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Brian's kite is flying above a field at the end of a 68m string.If the angle of elevation to the kite measures 70 degrees,how hi

Dvinal [7]

Answer:

$h = 63.92\ m$

Step-by-step explanation:

Given:

Angle of elevation = 70°

Length of string = 68 m

We need to find the height of the kite.

Solution:

Requirement figure attached in file.

Where:

BC = 68 m

AC = h (Height of the kite)

∠ABC = 70°

Using Cosine rule to find the height of the kite.

$AC = BC\times sin(70)$

$h = 68\times 0.94$

$h = 63.92\ m$

Therefore, height of the kite from Brian's head $h = 63.92\ m$ .

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Use fundamental theorem of calculus to find derivative of the function LOOK AT PHOTO

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By the FTC, the derivative is

$\displaystyle \frac{df}{dx} = \left(\frac1x + \sin\left(\frac1x\right)\right) \frac{d}{dx}\left[\frac1x\right] - (\ln(x) + \sin(\ln(x))) \frac{d}{dx}\left[\ln(x)\right] \\\\ = -\frac1{x^2} \left(\frac1x + \sin\left(\frac1x\right)\right) - \frac1x (\ln(x) + \sin(\ln(x))) \\\\ = -\frac1{x^3} - \frac{\sin\left(\frac1x\right)}{x^2} - \frac{\ln(x)}x - \frac{\sin(\ln(x))}x \\\\ = -\frac{1 + x\sin\left(\frac1x\right) + x^2\ln(x) + x^2 \sin(\ln(x))}{x^3}$

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