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mr_godi [17]
4 years ago
11

Estimate the intervals on which the derivative is positive and the intervals on which the derivative is negative. enter your ans

wers to one decimal place.
Mathematics
1 answer:
Norma-Jean [14]4 years ago
7 0
<span>Here is the general strategy for identifying whether a derivative is positive or negative. First Let f(x) be a function of x , continuous and differential on a given interval 'I'. f'(x) is positive if f(x) is increasing on 'I' . f'(x) is negative if f(x) is decreasing on 'I' .</span>
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A moving firm charges a flat fee of $35 plus $30 per hour. Let y be the cost in dollars of using the moving firm
zysi [14]

Answer:

y=35x+30

Step-by-step explanation:

7 0
3 years ago
Cual es la forma más simple al efectuar la operación indicada en la expresión (-3) (-12) ?
gulaghasi [49]

Answer:

Cual es la forma más simple al efectuar la operación indicada en la expresión (-3) (-12) ?  this means What is the simplest way to perform the operation indicated in the expression (-3) (-12)?

Step-by-step explanation:

Multiply -3 by -12 and get 36. hope this was helpful.

7 0
3 years ago
Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>​
Alla [95]

\large \mathbb{PROBLEM:}

\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}

\large \mathbb{SOLUTION:}

\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}

\boxed{ \tt   \red{C}arry  \: \red{ O}n \:  \red{L}earning}  \:  \underline{\tt{5/13/22}}

4 0
2 years ago
The length of a rectangle is shown below:
ludmilkaskok [199]

Answer:

C(5,-4), D(-4,-4)

Have a great day :)

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
Find c. Round to the nearest tenth.
mixer [17]

Answer:

Here,

150+10=180

162-180=18

Answer is 18°

Step-by-step explanation:

So the round of nearest tenth is 18°

Hope it will help have a great day at school. ^_^

3 0
3 years ago
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