Y varies inversely as the fourth power of x and when x=3, y=1.

Mathematics

1 answer:

masya89 [10]1 year ago

4 0

The constant of proportionality if y varies inversely as the fourth power of x and when x=3, y=1 is k = 3^¼

<h3>Inverse variation</h3>

y = k ÷ x^¼

where,

Constant of proportionality = k

When x = 3, y = 1

y = k ÷ x^¼

1 = k ÷ 3^¼

1 = k / 3^¼

cross product

1 × 3^¼ = k

k = 3^¼

Therefore, the constant of proportionality if y varies inversely as the fourth power of x and when x=3, y=1 is k = 3¼

Learn more about inverse variation:

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<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"

qwelly [4]

Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that $\arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)$

Also, $-\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)$

This means our original expression is equal to:

$\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]$

Using the cosine addition formula, which states $\cos(a+b)=\cos a \cos b-\sin a \sin b$ , we get this itself is equal to:

$\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)$

Since $\sin^{2} \theta+\cos^{2} \theta=1$ , we know that:

$\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}$

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

$\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}$