All categories

4 years ago

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

You might be interested in

Express this equation in logarithmic form. y = 10^x x = ?

Lana71 [14]

From definition we have:

$$x=\log_{10}y$

8 0

3 years ago

What is 3 ÷ 2\5 pls answer dis

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 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