20 points please help!!!!!

The area of the surface is 144.708

The equation of the given surface is,

z=g(x,y)=xy

Solving the partial derivatives,

∂g∂x=y,∂g∂y=x

Substituting to the formula

S=∬√1+( ∂g∂x)2+(∂g∂y)2dA

Thus,

S=∬√1+(y)2+(x)2dAS=∬√1+x2+y2dA

The region in the XY-plane is defined by the intervals  0≤θ≤2π,0≤r≤4

Converting the integral into polar coordinates,

S=∫2π0∫40√1+r2rdrdθ

Integrating with respect to r

S=∫2π0[13(1+r2)32]40dθ

S=∫2π0(17√173−13)dθ

Integrating with respect to θ

S=(17√173−13)[θ]2π0

S≈144.708

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Answer:

6

Step-by-step explanation:

45/8=5.625

Since there are .625 children left, we need to add another ride to fill in what's left.

Answer:

30 km/h

Step-by-step explanation:

Let X km/h be the speed of the boat in still water

Let the distance covered each way be D km

Relative speed of boat downstream = X + 6

Relative speed of boat upstream = X - 6

Distance covered by boat when going downstream

D = 4(X + 6) = 4X + 24

Distance covered by boat upstream is the same as distance covered downstream
D = 6(X-6) = 6X -36

Setting the two X equations equal to each other gives us

6X - 36 = 4X + 24

Collecting like terms

6X - 4X = 24-(-36)

2X = 60

X = 30 km/hr which is the speed of boat in still water

Check

Downstream speed is 30+6 = 36 km/h and distance traveled downstream in 4 hours = 36 x 4 = 144km

Upstream speed is 30-6 = 24 km/h and it travels for 6 hours. Distance covered = 24 x 6 = 144 miles

So distance covered is same and hence check succeeds

Answer:   9

Step-by-step explanation:

assuming that each box is 1 square inch then, you would get 9 square inches.

0,5 go down three to get 0,2

0,2 go left three to get -3,2

-3,2 go up three to get -3,5

-3,5 go right three to get 0,5

so you end up with a 3-by-3 box

Answer:

Work is incorrect

Step-by-step explanation:

I'm assuming this question is asking whether the work is correct or not? In which case the work is not correct.

The Pythagorean Theorem states: [tex]a^2+b^2=c^2[/tex] where c=hypotenuse, and "a" and "b" are the other two sides. The main thing to note here, is that the hypotenuse is the largest side of all three sides.

So the equation Arial set up: [tex]9^2+15^2=12^2[/tex] is incorrect, since the 15 would need to be on the right side. This forms the correct equation: [tex]9^2+12^2=15^2[/tex] which then simplifies to: [tex]81 + 144 =225 \implies 225=225[/tex].

Thus a right triangle can be formed using these side lengths. You can of course set up a similar equation to Ariels, where the "c" or hypotenuse is not isolated,  but you would have to rearrange the equation so that: [tex]a^2+b^2=c^2\implies b^2=c^2-a^2[/tex] but see how "a squared" is being subtracted from "c squared"? So it's a similar equation to Ariels, but not quite the same, and if she set it up like this, then she would reach the same conclussion

By just adding the given areas, we conclude that the definite integral is equal to 3.026

The integral will be equal to the sum of the four areas between points a and c.

The areas above the horizontal axis are positive areas, these are A and C.

While the areas below the horizontal axis are negative areas, these are B and D.

Then the definite integral will be:

I = A - B + C - D

Here we know that:

A = 1.311

B = 2.229

C =  5.545

D = 1.601

Replacing that, we conclude that the definite integral is

I =  1.311 - 2.229 +5.545 - 1.601 = 3.026

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This means that after 0.5 hours, 259.81 milligrams of the medicine remain on the patient's body.

We know that the exponential function defined by:

[tex]m(h) = 300*(3/4)^h[/tex]

Represents the amount of a mediciene, in milligrams, after a number of hours h.

We want to see what does m(0.5) represent. This is the exponential function evaluated in h = 0.5

Then, it just represents the amount of medicine, in miligrams, in a patient's body present 0.5 hours after the medicine was administered.

Replacing h = 0.5 we get:

[tex]m(0.5) = 300*(3/4)^{0.5} = 259.81[/tex]

This means that after 0.5 hours, 259.81 milligrams of the medicine remain on the patient's body.

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the gardener would need (1 + 1/6) liters of water to water the whole garden.

Here we know that the gardener needs 1/3 of a liter of water to water 2/7 of a garden.

Then we have the relation:

1/3 L = 2/7 of a garden.

Now, we want to get a "1 garden" in the right side of the equation, then we can multiply both sides by (7/2), so we get:

(7/2)*(1/3) L = (7/2)*(2/7) of a garden

(7/6)L = 1 garden.

(1 + 1/6) L = 1 garden

This means that the gardener would need (1 + 1/6) liters of water to water the whole garden.

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Answer:

40° to the left and 3/2 units down

Step-by-step explanation:

See attached image.

Answer:

Step-by-step explanation:

1144

Answer:

Step-by-step explanation:

The volume formula can be used to find the height and width of a box with volume 4368 in³ and height 1 in greater than width.

The volume formula is ...

  V = LWH

Substituting given information, using w for the width, we have ...

  4368 = (24)(w)(w+1)

We want to find the value of w.

  182 = w² +w . . . . . . . . divide by 24

  182.25 = w² +w +0.25 = (w +0.5)² . . . . . . add 0.25 to complete the square

  13.5 = w +0.5 . . . . . . . . take the positive square root

  w = 13 . . . . . . . . . . . . subtract 0.5

  h = w+1 = 14

The height of the box is 14 inches; the width is 13 inches.

__

Additional comment

By "completing the square", we can arrive at the exact dimensions of the box, as we did above. Note that we only added 0.25 to the equation to do this.

For numbers close together, the geometric mean (root of their product) is about the same as the arithmetic mean (half the sum):

  [tex]\sqrt{w(w+1)}\approx\dfrac{w+(w+1)}{2}=w+\dfrac{1}{2}\\\\w\approx\sqrt{182}-\dfrac{1}{2}\approx12.99[/tex]

Using this approximation to arrive at the conclusion w=13 saves the steps of figuring the value necessary to complete the square, then adding that before taking the root.

Answer:

Step-by-step explanation:

20 liters -1 liter = 19 liters of milk

20 liters -1 liter = 19 liters of syrup

This was done twice so therefore;

19 liters  -1 liter = 18 liters of milk

19 liters  -1 liter = 18 liters of syrup

Therefore the ratio comes to 18:18

18 goes into 18 1 and the same for the other 18. Therefore the ratio comes to ;

1:1

Your final answer is 1:1

Answer:

0.342

Step-by-step explanation:

round to the given amount of significant figures

Answer: [tex]\displaystyle\\\frac{x^3}{y^7}[/tex].

Step-by-step explanation:

[tex]\displaystyle\\x^3y^{-7}=\frac{x^3}{y^7} .[/tex]

Answer:

34

Step-by-step explanation:

(a + [tex]\frac{1}{a}[/tex] ) = 6← square both sides

(a + [tex]\frac{1}{a}[/tex] )² = 6² ← expand left side using FOIL

a² + 1 + 1 + [tex]\frac{1}{a^2}[/tex] = 36

a² + 2 + [tex]\frac{1}{a^2}[/tex] = 36 ( subtract 2 from both sides )

a² + [tex]\frac{1}{a^2}[/tex] = 34

Using it's concept, the average rate of change of the function on the interval [6,13] is given by:

[tex]r = \frac{f(13) - f(6)}{7}[/tex]

The average rate of change of a function is given by the change in the output divided by the change in the input. Hence, over an interval [a,b], the rate is given as follows:

[tex]r = \frac{f(b) - f(a)}{b - a}[/tex]

In this problem, we want to find the rate in the interval [6, 13], which is given as follows:

[tex]r = \frac{f(13) - f(6)}{13 - 6} = \frac{f(13) - f(6)}{7}[/tex]

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Answer:

56

Step-by-step explanation:

combinations

C(8,3)

Answer:

Option A

Step-by-step explanation:

The solution is in the image

Answer:

Rs 1425

Step-by-step explanation:

=(30×40) + (15×15)

=1425

Using the given data, the average percent score for the 100 students is 77[tex]\%[/tex].

Average, also known as mean, is one of the measures of central tendency and is the ratio of the sum of all observations to the total number of observations. It is very useful to summarize a probability distribution.

Let the average percent score for the 100 students be N.

As it is known that the average is the ratio of the sum of all observations to the total number of observations, therefore:

[tex]N = \dfrac{100\times7+90\times18+80\times35+70\times25+60\times10+50\times3+40\times2}{100}[/tex]

   [tex]= \dfrac{7700}{100}\\= 77[/tex]

Thus, the average percent score for the 100 students, using the given data is calculated as 77[tex]\%[/tex].

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The complete question is as follows:

Mrs. Riley recorded this information from a recent test taken by all of her students. Using the data, what was the average percent score for these 100 students?

[tex]\%[/tex] Score  Number of students

100                  7

90                   18

80                   35

70                   25

60                   10

50                   3

40                   2

the solution of the system of linear equations is:

x = 6/11

y = -4/11

Here we have the system of equations:

To solve the system, we can see that y is already isolated in both sides, then we get:

(-5/2)*x + 1 = 3x - 2

Now we can solve this for x:

(-5/2)*x + 1 = 3x - 2

2 + 1 = 3x + (5/2)*x

3 = (6/2)*x + (5/2)*x

3 = (11/2)*x

(2/11)*3 = x

6/11 = x

To get the value of y, we evaluate any of the lines in x = 6/11

y = 3*(6/11) - 2 = 18/11 - 2 = 18/11 - 22/11 = -4/11

Then the solution of the system of linear equations is:

x = 6/11

y = -4/11

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Answer:

9 or 16... but I'm leaning more to 9

Step-by-step explanation:

Since the graph seems to be going up and down, it seems like its at its highest point and should go down now so it could be 9. But if not and its still going to go higher 16 is possible.

Answer:

Allana is 2 1/2 and her mother is 15 years old.

Step-by-step explanation:

Let Allana's age be x years then,

her mother's age is 6x.

Using the given information in ten years time we have the equation:-

6x + 10 = 2(x + 10)

6x + 10 = 2x + 20

6x - 2x = 20 - 10

4x = 10

x = 2 1/2

So 6x = 6 * 2 1/2 = 15.

Answer:

[tex]\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}[/tex]

[tex]\textsf{Trapezium rule}: \quad \pi[/tex]

[tex]\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}[/tex]

Step-by-step explanation:

Midpoint rule

[tex]\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]

Trapezium rule

[tex]\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]

Simpson's rule

[tex]\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]

Given definite integral:

[tex]\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x[/tex]

Therefore:

Calculate the subdivisions:

[tex]\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi[/tex]

Midpoint rule

Sub-intervals are:

[tex]\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right][/tex]

The midpoints of these sub-intervals are:

[tex]\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi[/tex]

Therefore:

[tex]\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}[/tex]

Trapezium rule

[tex]\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}[/tex]

[tex]\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}[/tex]

Simpson's rule

[tex]\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}[/tex]

Answer:

Domain: (-∞, ∞), Range: [-3, ∞)

Step-by-step explanation:

Domain is all of the possible x-values in a solution set, in this case since the solution extends infinitely horizontally it would be:

(-∞, ∞)

The range is just all of the possible y-values in a solution set, which in this set starts from -3 and proceeds on infinitely, making the range:

[-3, ∞)

Answer:

406

Step-by-step explanation:

let the 5 consecutive positive integers be

n , n + 1 , n + 2 , n + 3 , n + 4 , then

n + n + 1 + n + 2 + n + 3 + n + 4 = 2020

5n + 10 = 2020 ( subtract 10 from both sides )

5n = 2010 ( divide both sides by 5 )

n = 402

then

largest integer = n + 4 = 402 + 4 = 406

The number line which represents the solution bro the inequality y -2 < -5 is;

An open circle is at negative 3. Everything to the left of the circle is shaded.

First, we must solve the inequality so that it looks it's simplest form in order to be represented on a number line.

Solving the inequality is as follows;

y - 2 < -5

y < -5 +2

y < -3.

In essence, representation of the inequality, y < -3 on the number line is as follows;

An open circle is at negative 3. Everything to the left of the circle is shaded.

PS: It is only a closed circle if the inequality includes an equal to sign.

It would be the bottom left graph

Initial production: 1050

so the graph will be started at 1050 which makes top right graph incorrect

Next, the money at the lowest was 250 which makes the top left and bottom right incorrect.

leaving only the bottom right graph to be correct

Answer:

Graph Y

Step-by-step explanation:

Given information:

The initial production cost is when the number of items is zero.

Therefore, the y-intercept of the graph will be (0, 1050).

The lowest production cost is the minimum point of the curve.

Therefore, the vertex of the graph will be (2, 250).

The only graph that satisfies these conditions is graph Y (attached).

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