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3 years ago

Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

7 0

The answer is 100.53.

This is the circumference using the formula C=2πr .

ziro4ka [17]3 years ago

6 0

Answer:

C = 32π

Step-by-step explanation:

The circumference (C) of a circle is calculated as

C = 2πr ( r is the radius ) , then

C = 2π × 16 = 32π units

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Pls help me with this question!!!

yarga [219]

<h3>In which language u will speak...</h3>

6 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: <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>

5 0

4 years ago

1) Find the minimum and maximum values for the function with the given domain interval.

Julli [10]

Answer:

"minimum value = 0; maximum value = 8"

Step-by-step explanation:

This is the absolute value function, which returns a positive value for any numbers (positive or negative).

For example,

| -9 | = 9

| 9 | = 9

| 0 | = 0

Now, the domain is from -8 to 7 and we want to find max and min value that we can get from this function.

If we look closely, putting 7 into x won't give us max value as putting -8 would do, because:

|7| = 7

|-8| = 8

So, putting -8 would give us max value of 8 for the function.

Now, we can't get any min values that are negative, because the function doesn't return any negative values. So the lowest value would definitely be 0!

|0| = 0

and

ex: |-2| = 2 (bigger), |-5| = 5 (even bigger).

So,

Min Value = 0

Max Value = 8

8 0

3 years ago

Read 2 more answers

A florist delivers flowers to anywhere in town. D is the distance from the delivery address to the florist shop in miles. The co

Inessa [10]

A. C (6) = 15.24; C 11 = 22.09.

C (6) represents the cost, $15.24, of delivering flowers to a destination that is 6 miles

from the shop.

C (11) represents the cost, $22.09, of delivering flowers to a destination that is 11 miles

from the shop.

8 0

3 years ago

Read 2 more answers

Verdich [7]

Answer:

A. Please see the attached graphs

B. 5. The equation of the graph is y = 3·x - 3

6. The equation of the graph is y = x + 3

7. The equation of the graph is y = 4/5·x - 4

8. The equation of the graph is y = 2·x - 4

9. The graph is the line with equation y = 5·x - 31

10. The graph is the line with equation y = 5·x - 14

11. The graph is the line with equation y = -2·x + 9

12. The graph is the line with equation y = 3·x + 6

Step-by-step explanation:

A. Please see the attached graphs

B. 5. The intercepts are;

(0, -3) and (1, 0)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (0, -3) and (1, 0) we have;

m = (0 - (-3))/(1 - 0) = 3

The slope = 3

From the point and slope form, of a straight line equation, we have;

y - 0 = 3(x - 1)

The equation of the graph is therefore;

y = 3·x - 3

The y-intercept occurs at (0, -3)

The x intercept occurs where y = 0

0 = 3·x - 3

x = 3/3 = 1

The x-intercept occurs at (1, 0)

The graph of the equation, y = 3·x - 3, passes through the y and x intercepts (0, -3) and (1, 0) respectively

6. The coordinate points are;

(-3, 0) and (0, -3)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (-3, 0) and (0, -3) we have;

m = ((-3) - 0)/(0 - (-3)) = -1

The slope = 1

From the point and slope form, of a straight line equation, we have;

y - 0 = 1(x - (-3)) = x + 3

y = x + 3

The y-intercept occurs at (0, 3)

The x intercept occurs where y = 0

0 = x + 3

x = -3

The x-intercept occurs at (-3, 0)

The graph of the equation, y = x + 3, passes through the y and x intercepts (0, 3) and (-3, 0) respectively

7. The coordinate points are;

(5, 0) and (0, -4)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (5, 0) and (0, -4) we have;

m = ((-4) - 0)/(0 - 5) = 4/5

The slope = 4/5

From the point and slope form, of a straight line equation, we have;

y - 0 = 4/5×(x - 5) = -4/5·x - 4

y = 4/5·x - 4

The y-intercept occurs at (0, -4)

The x intercept occurs where y = 0

0 = -4/5·x + 4

-4 = -4/5·x

x = 5

The x-intercept occurs at (5, 0)

The graph of the equation, y = 4/5·x - 4, passes through the y and x intercept (0, 4) and (5, 0) respectively

8. The coordinate points are;

(2, 0) and (0, -4)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (2, 0) and (0, -4) we have;

m = ((-4) - 0)/(0 - 2) = 2

The slope =2

From the point and slope form, of a straight line equation, we have;

y - 0 = 2×(x - 2) = 2·x - 4

y = 2·x - 4

The y-intercept occurs at (0, -4)

The x intercept occurs where y = 0

0 = 2·x - 4

4 = 2·x

x = 2

The x-intercept occurs at (2, 0)

The graph of the equation, y = 2·x - 4 passes through the y and x intercept (0, -4) and (2, 0) respectively

C. Using the point and slope form

9. The slope m = 5 and the graph passes through the points (6, -1)

We have the point and slope form given as follows;

y - (-1) = 5·(x - 6)

y = 5·x - 30 - 1 = 5·x - 31

y = 5·x - 31

The graph is the line with equation y = 5·x - 31

10. The slope m = 5 and the graph passes through the points (2, -4)

We have the point and slope form given as follows;

y - (-4) = 5·(x - 2)

y = 5·x - 10 - 4 = 5·x - 14

y = 5·x - 14

The graph is the line with equation y = 5·x - 14

11. The slope m = -2 and the graph passes through the points (4, 1)

We have the point and slope form given as follows;

y - 1 = (-2)·(x - 4)

y = -2·x + 8 + 1 = -2·x + 9

y = -2·x + 9

The graph is the line with equation y = -2·x + 9

12. The slope m = 3 and the graph passes through the points (-3, -3)

We have the point and slope form given as follows;

y - (-3) = 3·(x - (-3))

y = 3·x + 9 - 3 = 3·x + 6

y = 3·x + 6

The graph is the line with equation y = 3·x + 6

3 0

3 years ago