Question

3. solve the following word problems b) The state government announced an award of 75,000 for each of the 18
medal winners of the recently held international games. How much money
will the government be disbursing?​

Answer 1

Answer:

5.minute.crafts is a very important person for the next couple more friends but the other two people.

Step-by-step explanation:

• 1

1. 2

• 3

1. 4

• 5

1. 6

• 7

1. 8

• 9

1. 10

Answer 2

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Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x

Answer 3

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Topic :-This is your punishment for spamming in my question.

If you do not know the answer, refrain from answering.

Don't be greedy for points.

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x

Answer 4

$\huge \: \fbox \red {{{ \colorbox{green}{{answer}}}}}$

Topic :-This is your punishment for spamming in my question.

If you do not know the answer, refrain from answering.

Don't be greedy for points.

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x

Answer 5

$\huge \: \fbox \red {{{ \colorbox{green}{{answer}}}}}$

Topic :-This is your punishment for spamming in my question.

If you do not know the answer, refrain from answering.

Don't be greedy for points.

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x

Answer 6

$\huge \: \fbox \red {{{ \colorbox{green}{{answer}}}}}$

Topic :-This is your punishment for spamming in my question.

If you do not know the answer, refrain from answering.

Don't be greedy for points.

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x

Answer 7

$\huge \: \fbox \red {{{ \colorbox{green}{{answer}}}}}$

Topic :-This is your punishment for spamming in my question.

If you do not know the answer, refrain from answering.

Don't be greedy for points.

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x