All categories

2 years ago

What was the purpose of the townshend program

Mathematics

1 answer:

kykrilka [37]2 years ago

4 0

This was to raise revenue for the British Empire by taxing the North American colonies.

You might be interested in

Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$

Nothing to do to simplify any further, other than factoring out $x^2$ .

4 0

3 years ago

Describe a real-word example of when it is necessary that line segments are parallel

Alex_Xolod [135]

If you are drawing a map, you may need to draw two parallel line segments.

8 0

3 years ago

Read 2 more answers

PLEASE PLEASE PLEASE PLEASE HELP!! even if it's only one! actually answer it!

Alex

So for number 2 i am 100% on which the area of the triangle is 6.
and then 1 i think not 100% sure is an acute angle again not 100%
then 3. it don't make sense to me so idk about that 1 srry.
~Good Luck~

6 0

3 years ago

I need help with this I got -10 is it right??

Dahasolnce [82]

Answer:

-14

Step-by-step explanation:

difference between -20 and -5 = 15

15 x 2/5 = 30/5 = 6

-20 + 6 = -14

7 0

3 years ago

5 questions plz answer all of them if you do you will get points, thanks, Brainliest answer,

babunello [35]

1. 1/6

2. 1/8

3. 80%

4. 1/5

5. 2/8 or 25%

5 0

3 years ago