Find each interior angles the following regular polygons in degrees and grades
(a) Triangle
(b) Quadrilateral (c) Pentagon (d) Hexagon
(e) Heptagon
(1) Octagon (g) Nonagon
(h) Decagon
polygons in degrees and grades​

Answers

Answer 1
Answer:

(a) 120° or 133.33 grades.

(b)135° or 150 grades.

(c) 144° or 160 grades.

(d) 150° or 166.67 grades.

(e) 154.29° or 171.43 grades.

(f) 157.5° or 175 grades.

(g) 160° or 177.78 grades.

(h) 162° or 180 grades.

Step-by-step explanation:

A regular polygon is a polygon that has all sides and all angles equal.

Each interior angle, k (measured in degrees), of a regular polygon is given by;

k = s ÷ n            -----------(k)

Where;

s = sum of the interior angle of the polygon.

n = number of sides of the polygon

To get s, we use

s = (n - 2) x 180         [This is the formula to calculate the sum of the interior angles of a polygon]

Substituting s this into equation (i) gives

k = (n - 2) x 180 ÷ n

k = 180(n - 2) ÷ n             -------------(ii)

(a) Triangle.

A triangle has n = 3 sides, therefore, each interior angle of a triangle is found by substituting n = 3 into equation (ii)

k = 180(3 - 1) ÷ 3

k = 180(2) ÷ 3

k = 180 x 2 ÷ 3

k = 360 ÷ 3

k = 120°

Convert to grade.

Remember that;

90° = 100 grades

∴ 120° = [tex]\frac{120 * 100}{90}[/tex] grades

⇒ 120° = 133.33 grades.

Therefore, each interior angles of a regular triangle is 120° or 133.33 grades.

(b) Quadrilateral.

A quadrilateral has n = 4 sides, therefore, each interior angle of a quadrilateral is found by substituting n = 4 into equation (ii)

k = 180(4 - 1) ÷ 4

k = 180(3) ÷ 4

k = 180 x 3 ÷ 4

k = 540 ÷ 4

k = 135°

Convert to grade.

Remember that;

90° = 100 grades

∴ 135° = [tex]\frac{135 * 100}{90}[/tex] grades

⇒ 135° = 150 grades.

Therefore, each interior angles of a regular quadrilateral is 135° or 150 grades.

(c) Pentagon

A pentagon has n = 5 sides, therefore, each interior angle of a pentagon is found by substituting n = 5 into equation (ii)

k = 180(5 - 1) ÷ 5

k = 180(4) ÷ 5

k = 180 x 4 ÷ 5

k = 720 ÷ 5

k = 144°

Convert to grade.

Remember that;

90° = 100 grades

∴ 144° = [tex]\frac{144 * 100}{90}[/tex] grades

⇒ 144° = 160 grades.

Therefore, each interior angles of a regular pentagon is 144° or 160 grades.

(d) Hexagon

A hexagon has n = 6 sides, therefore, each interior angle of a hexagon is found by substituting n = 6 into equation (ii)

k = 180(6 - 1) ÷ 6

k = 180(5) ÷ 6

k = 180 x 5 ÷ 6

k = 900 ÷ 6

k = 150°

Convert to grade.

Remember that;

90° = 100 grades

∴ 150° = [tex]\frac{150 * 100}{90}[/tex] grades

⇒ 150° = 166.67 grades.

Therefore, each interior angles of a regular hexagon is 150° or 166.67 grades.

(e) Heptagon

A heptagon has n = 7 sides, therefore, each interior angle of a heptagon is found by substituting n = 7 into equation (ii)

k = 180(7 - 1) ÷ 7

k = 180(6) ÷ 7

k = 180 x 6 ÷ 7

k = 1080 ÷ 7

k = 154.29°

Convert to grade.

Remember that;

90° = 100 grades

∴ 154.29° = [tex]\frac{154.29 * 100}{90}[/tex] grades

⇒ 154.29° = 171.43 grades.

Therefore, each interior angles of a regular heptagon is 154.29° or 171.43 grades.

(f) Octagon

An octagon has n = 8  sides, therefore, each interior angle of a octagon is found by substituting n = 8 into equation (ii)

k = 180(8 - 1) ÷ 8

k = 180(7) ÷ 8

k = 180 x 7 ÷ 8

k = 1260 ÷ 8

k = 157.5°

Convert to grade.

Remember that;

90° = 100 grades

∴ 157.5° = [tex]\frac{157.5 * 100}{90}[/tex] grades

⇒ 157.5° = 175 grades.

Therefore, each interior angles of a regular octagon is 157.5° or 175 grades.

(g) Nonagon

A nonagon has n = 9 sides, therefore, each interior angle of a nonagon is found by substituting n = 9 into equation (ii)

k = 180(9 - 1) ÷ 9

k = 180(8) ÷ 9

k = 180 x 8 ÷ 9

k = 1440 ÷ 9

k = 160°

Convert to grade.

Remember that;

90° = 100 grades

∴ 160° = [tex]\frac{160 * 100}{90}[/tex] grades

⇒ 160° = 177.78 grades.

Therefore, each interior angles of a regular nonagon is 160° or 177.78 grades.

(h) Decagon

A decagon has n = 10 sides, therefore, each interior angle of a decagon is found by substituting n = 10 into equation (ii)

k = 180(10 - 1) ÷ 10

k = 180(9) ÷ 10

k = 180 x 9 ÷ 10

k = 1620 ÷ 10

k = 162°

Convert to grade.

Remember that;

90° = 100 grades

∴ 162° = [tex]\frac{162 * 100}{90}[/tex] grades

⇒ 162° = 180 grades.

Therefore, each interior angles of a regular decagon is 162° or 180 grades.


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Answers

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Answers

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Answers

Answer:

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Answers

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Answers

Answer:

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Answers

Answer:

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Answers

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Answer:

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Answers

Answer:

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Answers

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10.63

8.)

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10.58

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1.) because 10^3=1000 and 11^3 = around 1330  we know that the answer must fall between 10 and 11. Because they're pretty much equal distance I would estimate it to be around 10.5 (if you need an exact answer it's 2∛150)

2.) because 11^2=121 and 10^2= 100 we know that it must fall between 10 and 11. Because these are pretty much equal distance as well I would estimate it to be around 10.5 as well (if you need an exact answer it's 4√7)

Step-by-step explanation:

6 Factorise
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Answers

The answer should be (x-7) (x+1)

Answer:

(x+7)(x-1)

Step-by-step explanation:

x^2 + 6x -7

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7 and 1 are the numbers because

7*1=7  and ,

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x^2 + (7-1)x -7

x^2 +7x -x -7

take common from first two terms and from third and fourth term

x(x+7)-1(x+7)

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(x+7)(x*1 -1*1)

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Answers

Answer:

Amount won : $3,000,000

Step-by-step explanation:

Let the amount won be =100 x

Amount spent on house :

                                     [tex]\frac{4}{5} \ of \ 100 x = 80 x[/tex]

Remaining amount = 100x - 80x = 20x

Amount spent on yacht :

                                    [tex]\frac{3}{4} \ of \ 20x = 15x[/tex]

Remaining amount = 20x - 15x = 5x

Amount spent on a trip :

                                   [tex]\frac{2}{3} \ of \ 5x = \frac{10}{3}x[/tex]

Remaining amount =

                         [tex]5x - \frac{10}{3}x = \frac{15x -10x}{3} = \frac{5x}{3}[/tex]

This amount is given for charity :

                                                     [tex]\frac{5}{3}x[/tex]

But given :

           Amount given to charity is  50000

Therefore ,

                 [tex]\frac{5}{3}x = 50000[/tex]

                [tex]5x = 3 \times 50000\\\\5x = 150000\\\\x = 30000[/tex]

Therefore , the amount won on lottery is = 100x = 100 ( 30000) = 3,000,000

The amount of money he won

A student's course grade is based on one midterm that counts as 20% of his final grade one class project that counts as 20% of his final grade a set of homework assignments that counts as 30% of his final grade, and a final exam
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His overall final score is
(Round to one decimal place as needed)

Answers

the overall score is 86.7

The total body surface area, or BSA, of a human is difficult to calculate. There are various models that estimate BSA based on a person's weight and height. One simpler model is BSA= wh 3600
where w = weight in kg and h = height in cm.
a. Using this model, estimate the height of a person who weighs 75 kg and whose BSA is 1.9.
Round your answer to the nearest cm.
h = _______ cm
b. Using this model, estimate the weight of a person who is 166 cm tall and whose BSA is 2.2.
Round your answer to the nearest kg.
w= _________kg

Answers

Answer:

173 cm

105 kg

Step-by-step explanation:

Given :

BSA= √wh /3600

where w = weight in kg and h = height in cm.

A.)

w = 75 kg ; BSA = 1.9 ; h =?

BSA= √wh /3600

1.9 = √75h /3600

Take the square of both sides

1.9² = 75h /3600

1.9² * 3600 = 75h

12996 = 75h

h = 12996 / 75

h = 173.28 cm

h = 173 cm

B.)

h = 166 ; BSA = 2.2

BSA= √wh /3600

2.2 = √w166 /3600

Take the square of both sides

2.2² = 166w /3600

2.2² * 3600 = 166w

17424 = 166w

w = 17424 / 166

w = 104.96 kg

w = 105 kg

Using the model given, we have that:

a) h = 173 cm.

b) w = 105 kg.

The BSA, for a person of weight w and height h, is given by:

[tex]BSA = \sqrt{\frac{wh}{3600}}[/tex]

Item a:

[tex]BSA = 1.9, w = 75[/tex], hence:

[tex]BSA = \sqrt{\frac{wh}{3600}}[/tex]

[tex]1.9 = \sqrt{\frac{75h}{3600}}[/tex]

[tex]\left(\sqrt{\frac{75h}{3600}}\right)^2 = 1.9^2[/tex]

[tex]\frac{75h}{3600} = 3.61[/tex]

[tex]h = \frac{3600(3.61)}{75}[/tex]

[tex]h = 173[/tex]

Hence:

h = 173 cm.

Item b:

[tex]BSA = 2.2, h = 166[/tex], hence:

[tex]BSA = \sqrt{\frac{wh}{3600}}[/tex]

[tex]2.2 = \sqrt{\frac{166w}{3600}}[/tex]

[tex]\left(\sqrt{\frac{166w}{3600}}\right)^2 = 2.2^2[/tex]

[tex]\frac{166w}{3600} = 4.84[/tex]

[tex]w = \frac{3600(4.84)}{166}[/tex]

[tex]w = 105[/tex]

Hence:

w = 105 kg.

A similar problem, in which there is also an application of a formula, is given at https://brainly.com/question/24348510

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