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LiRa [457]
4 years ago
9

If a question asks you to find f(2) ,what does that mean?

Mathematics
1 answer:
Effectus [21]4 years ago
7 0
Plug 2 into the x in the equation
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Express this equation in logarithmic form. y = 10^x<br><br> x = ?
Lana71 [14]

From definition we have:

$x=\log_{10}y

8 0
3 years ago
What is 3 ÷ 2\5 pls answer dis<br>​
Anna35 [415]

Answer:

15/2

Step-by-step explanation:

when dividing by fractions, we "flip and multiply"-- therefore:

3 ÷ 2\5 = 3 x 5/2

3 x 5/2 = 15/2

7 0
3 years ago
A block of gold with a mass of 3864 kilograms has a volume of 0.2 cubic meters. What is the density of gold?
kaheart [24]
The density of the gold is 772.8 kilograms\cubic meters
3 0
3 years ago
Solve for x<br><br> Need help with this last question
Serhud [2]

Answer:

x = 8

Step-by-step explanation:

2(78°) = 20x - 4

156 = 20x - 4

add 4 to each side of the equation:

160 = 20x

divide both sides by 20:

x = 8

6 0
3 years ago
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}

4 0
3 years ago
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