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3 years ago

If cos θ =-2/5 and tan θ > 0, what is the value of sin θ?

1 answer:

8 0

If the value of cos(θ) is negative, this angle (θ) can only be in one of two quadrants; the sine quadrant or the tan quadrant. In the sine quadrant, tan(θ) must be negative, but since tan(θ) > 0, we can safely say that the angle (θ) is based in the tan quadrant.

We know that cos(θ) = - Adjacent / Hypotenuse, and in this case Adjacent = 2 and Hypotenuse = 5. Using Pythagoras' theorem, we can find the opposite side of the right angled triangle situated in the tan quadrant...

Adjacent² + Opposite² = Hypotenuse²

Therefore:

2² + Opposite² = 5²

Opposite² = 5² - 2²

Opposite² = 21

Opposite = √(21)

------------

Now, sin(θ) must be negative, as the right angled triangle is in the tan quadrant. We also know that sin(θ) = Opposite / Hypotenuse, therefore:

sin(θ) = - [√(21)]/[5]

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Find the differential coefficient of <img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

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LiRa [457]

Answer:

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Step-by-step explanation:

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What’s the proof for this

KIM [24]

Well... Basically, you should prove this by SSS property(side-side-side). It's fair to say that the length of a side is equal to itself so The line that cuts through the rectangle is a side for both rectangles. Thus because of the given, all the sides of the triangles are equal to one another. This is a very important trick for geometry(I remember using it a lot).
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3 years ago

Simplifying square roots Please help!

seraphim [82]

I showed my work and answer in this picture. When simplifying square roots you can split them into a perfect squares root time the remainders