What is cosθ = _____ based on this image a)-3/5 b)-4/3 c)-4/5 d)-3/4

You have 16 digs for your current volleyball season. There are 3 games left in the season. You want to break your previous recor

natulia [17]

Answer: 16+d>20

d>4

Step-by-step explanation:

d=digs

16+d>20

16-16+d>20-16

d>4

7 0

1 year ago

At a shelter, 15% of the dogs are puppies. there are 60 dogs at the shelter. How many are puppies?

frutty [35]

So for this you need to find 15% of 60. There are a few ways to do this but I prefer to do it like this.

First, find 10% of 60. This can be found by dividing 60 by 10, because 100%/10% is 10.

This gives 10%=6

5% is half of 10%, so divide your answer for 10% (6), by 2.

This gives 5%=3

We needed 15%, so we add our 10% and 5% together, to get 9.

15% of 60 is 9, meaning there are 9 puppies.

Hope this helps!

8 0

3 years ago

Read 2 more answers

Please answer the question

maw [93]

The answer would be 208 if you find the area of each quadrilateral and add them together.

6 0

3 years ago

Read 2 more answers

A television with a 5:3 screen shows an image with a ratio of 5:4 which creates a letter boxed image. What percent of the screen

lions [1.4K]

Answer:

C

Step-by-step explanation:

8 0

3 years ago

∫(cosx) / (sin²x) dx

kirza4 [7]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2822772

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\
=\mathsf{\displaystyle\int\! \frac{1}{(sin\,x)^2}\cdot cos\,x\,dx\qquad\quad(i)}$

Make the following substitution:

$\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}$

and then, the integral (i) becomes

$=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
=\mathsf{\displaystyle\int\! u^{-2}\,du}$

Integrate it by applying the power rule:

$\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\
\mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\
\mathsf{=-\,\dfrac{1}{u}+C}$

Now, substitute back for u = sin x, so the result is given in terms of x:

$\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\
\mathsf{=-\,csc\,x+C}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus

5 0

3 years ago