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3 years ago

How do i turn 2/5 into a percentage

Mathematics

1 answer:

likoan [24]3 years ago

7 0

Numerator divided by the denominator then multiplied by 100.

(2/5) * 100

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Rewrite with only sin x and cos x.

Annette [7]

Option A

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

Solution:

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

$cos 3x = cos(x + 2x)$

We know that,

$\cos (a+b)=\cos a \cos b-\sin a \sin b$

Therefore,

$\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x$ ---- eqn 1

We know that,

$\sin 2 x=2 \sin x \cos x$

$\cos 2 x=\cos ^{2} x-\sin ^{2} x$

Substituting these values in eqn 1

$\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)$ -------- eqn 2

We know that,

$\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x$

Applying this in above eqn 2, we get

$\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)$

$\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}$

$\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x$

Therefore,

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

Option A is correct

7 0

3 years ago

Find the 2nd Derivative: f(x) = 3x⁴ + 2x² - 8x + 4

ad-work [718]

Answer:

$f''(x)=36x^2+4$

Step-by-step explanation:

Let's start by finding the first derivative of $f(x)= 3x^4+2x^2-8x+4$ . We can do so by using the power rule for derivatives.

The power rule states that:

$\frac{d}{dx} (x^n) = n \times x^n^-^1$

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

$\frac{d}{dx} (3x^4+2x^2-8x+4)$

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)$

Simplify the equation.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)$
$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)$

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)$
$f'(x)=12x^3+4x-8$

Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).