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2 years ago

6

Find the length of side x in simplest radical form with a rational denominator.

1 answer:

mojhsa [17]2 years ago

6 0

Answer:

x = $\frac{8\sqrt{3} }{3}$

Step-by-step explanation:

Using the sine ratio in the right triangle and the exact value

sin60° = $\frac{\sqrt{3} }{2}$

sin60° = $\frac{opposite}{hypotenuse}$ = $\frac{4}{x}$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

x × $\sqrt{3}$ = 8 ( divide both sides by $\sqrt{3}$ )

x = $\frac{8}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$ ( rationalising the denominator )

x = $\frac{8\sqrt{3} }{3}$

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dlinn [17]

Answer:

Statement 3

Step-by-step explanation:

y = -2/3x-24

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x = -(3/2)(y + 24)

Intersection variables

f^-1(x) = -(3/2)(x + 24)

Slope: -3/2

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3 years ago

How many people are added to the world population each year answer?

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2 years ago

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What times 9 equal 24

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3 years ago

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Find F"(x) if f(x) = cot (x)

hammer [34]

$f(x)=\cot x\implies f'(x)=-\csc^2x\implies\boxed{f''(x)=2\csc^2x\cot x}$

If you don't know the first derivative of $\cot$ , but you do for $\sin$ and $\cos$ , you can derive the former via the quotient rule:

$\cot x=\dfrac{\cos x}{\sin x}$

$\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x$

or if you know the derivative of $\tan$ :

$\cot x=\dfrac1{\tan x}$

$\implies(\cot x)'=-(\tan x)^{-2}\sec^2x=-\dfrac{\sec^2x}{\tan^2x}=-\dfrac{\frac1{\cos^2x}}{\frac{\sin^2x}{\cos^2x}}=-\dfrac1{\sin^2x}=-\csc^2x$

As for the second derivative, you can use the power/chain rules:

$(-\csc^2x)'=-2\csc x(\csc x)'=-2\csc x(-\csc x\cot x)=2\csc^2x\cot x$

or if you don't know the derivative of $\csc$ ,

$\csc x=\dfrac1{\sin x}$

$\implies(-\csc^2x)'=\left(-(\sin x)^{-2}\right)'=2(\sin x)^{-3}(\sin x)'=\dfrac{2\cos x}{\sin^3x}$

which is the same as the previous result since

$\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}$

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2 years ago

Pythagorean's theorem<br> 16^2+b^2=34^2

umka2103 [35]

Answer: 30

Step-by-step explanation:

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8 0

2 years ago