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sukhopar [10]
3 years ago
12

What is 20 5/81 in simplest form

Mathematics
1 answer:
Marizza181 [45]3 years ago
5 0
20 5/81 in its simplest form is 20 5/81.

20 5/81 originally looks like this ⇒ 1625/81

((20 * 81)+5)/81 = (1620+5)/81 = 1625/81   * this is an improper fraction because the numerator is greater than the denominator. Improper fractions must be converted to proper fractions. It will then become a mixed fraction.
 
     <u>  0020  5/81</u>        
81 | 1625
       <u>162      </u>      * 81 x 2
           05

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37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
Help plz ..............
Artemon [7]
First I would find the base, which is a square so 10x10=100
then find the triangle (a=1/2bh) a=1/2(12)(10) a=1/2(120) a=60
since there are four triangle sides do 60x4 which equals 240, so  then add the sides to the base 100+240=340, the answer is A
3 0
3 years ago
Find the volume of the cylinder. Round your answer to the nearest tenth. Use 3.14 for 1.
uranmaximum [27]

Answer:

791.28 in^2

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h

The diameter is 12 so the radius is 1/2 d = 1/2(12) =6

V = 3.14 (6)^2 * 7

791.28 in^2

5 0
3 years ago
Can someone please help me with this assignment
KIM [24]
These are 8 questions and 8 answers.

Page 1:

1) Triangle ABC has vertices A(-2,1), B(0,5), and C (8,1). What type of triangle is triangle ABC


Answer: option A. acute triangle.

Explanation:

Determine the lengths of the three sides:

Side AB: √[(0+2)^2 + (5-1)^2 ] = √ [4 + 16] = √(20)

Side BC: √ [ (8-0)^2 + (1 -5)^2 ] = √ [ 64 + 16 ] = √(86)

Side AC: √ [ (8 + 2)^2 + (1 - 1)^2 ] = √100 = 10

Compare the square of the side length of the largest side with the sum of the squares of the other two sides.:

AC^2 = 100
BC^2 = 86
AB^2 = 20

BC^2 + AB^2 = 86 + 20 = 106

100 < 106 => AC^2 < BC^2 + AB^2 =>

all the angles are less than 90°, which means the triangle is actue.

That is becasue when the square of the largest side is equal to the sum of the squares of the smaller sides, there is a right angle and the triangle is a right triangle (Pytaghorean theorem). When the square of the largest side is less than the sum of the squares of the other two sides, the angle is less than 90°, and the triangle is acute.

2) Which set represents the side lengths of a right trianle?

Answer: option C: 9, 12, 15

Explanation:

Check whic set obeys Pythagorean theorem.

15^2 = 225
12^2 = 144
9^2 = 81
.
144 + 81 = 225 => 15^2 = 12^2 + 9^2 => right triangle

3) Which property disproves the statement in the student's notes?

Answer: option B) all angles of a rhombus are not congruent.

Explanation:

A rhombus is a quadrilateral whose four sides all have the same length. It is a quadrilateral.

The angles do not need to be congruent. Only the square, which is only one special case of rhombus, has the four angles congruent, but the masure of the angles of other rhombus are not equal.

4) What is the are of the figure?

Answer: 85 unit^2

Explanation:

1) Area of the rectangle = base * height  = 4 * 9 = 36
2) area of the trapezoid = height * half of the sum of the bases = 7 * (3+7)/2 = 35
3) area of the rhombus = half the product of the diagonals = 7 * 4 / 2 = 14

Result: 36 + 35 + 14 = 85

That option is not among the choices. The nearest one is 84 unit^2.

Page 2 is the same page 1.

Page 3:

5) Question 8:

What line can be used to draw the altitude of the triangle FGH

Answer: option D: 3y = 7x + 16

Explanation:

1) All triangles have 3 altitudes (one for each vertex).

2) Now I am going to find the altitude of the vertex H (respect to base FG.

3) First calculate the slope of the segment FG:

slope = [y2 - y1] / [x2 - x1]

x2, y2 are the coordinates of the point G (6, -2) and x1, y1 are the coordinates of the point F (-8,4)

=> slope = [ -2 - 4] / [6 - (-8) ] = [-6] / [14] = - 3/7

4) Now calculate the slope of the perpendicular line (because the altitude is perpendicular to the base).

The lines of perpendicular lines obey the rule: slope line A = - slope line B / 2

So, the slope of the line perpendicular to the base is - [1 / (-3/7) ] = 7/3

5) The equation of the line with slope 3/7 that contains the point H (-4, - 4) is:

y - (-4)        7
--------- =  -----
x - (-4)       3

=> 3(y + 4) = 7(x + 4)

=> 3y + 12 = 7x + 28

=> 3y = 7x + 28 - 12

=> 3y = 7x + 16 which is the answer

6) Question 8. How will be the volume affected

Answer: option c. the volume will be 9 times larger.

Explanation:

1) Formula of the volume of rectagular prism

Volume = length * width * height

2) Name, V the volume, l the length, w the width, h the heigth => V = l * w * h

3) Triple the length and the width:

New volume = 3l * 3w * h = 9 l * w * h = 9V

4) You have proved that the new volumen is 9 times the original volume, which is the option c.


7) Question 9

What is the approximate length of the median of the trapezoid ABCD.

Answer: option b. 7.8 units

Explanation:

1) The median of a trapezoid is half the sum of the two parallel segments

2) The two parallel segments are AB and CD

3) The length of the segment AB is:

AB^2 = (3+7)^2 + (7-2)^2 = 10^2 + 5^2 = 100 + 25 = 125

=> AB = √125 ≈ 11.18

4) The length of the sement CD is:

CD^2 = (3 + 1)^2 + (1+1)^2 = 4^2 + 2^2 = 16 + 4 = 20

=> CD = √20 ≈ 4.47

5) The median is [11.48 + 4.47] / 2 ≈ 7.83 => option b.

8) Question 10. Problem

Answer: option B. 34.8 feet

Explanation:

1) Variables:

Ha = 4 ft (heigth of Abigail)
α = 38°
s = 50 ft (length of the string)
Hk = ? (altitude of the kite)

2) formula of sine ratio:

sin α = opposite-leg / adjacent leg

3) solution

sin α = x / s => x = sinα * s = sin(38°) * 50 ft = 30.78 ft

Hk = x + Ha = 30.78 ft + 4 ft = 34.78 ft => aption B. 34.8 feet
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3 years ago
The number of students in the 8 schools in a district are given below(Note that these are already ordered from least to greatest
vredina [299]
It would be 53 less then
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4 years ago
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