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Sholpan [36]
3 years ago
7

Find perimeter of regular pentagon is 100 inches. Find the length of each side.

Mathematics
2 answers:
RideAnS [48]3 years ago
8 0
<h2>Answer:</h2>

The length of each side of the polygon is:

                        20 inches.

<h2>Step-by-step explanation:</h2>

<u>Regular polygon--</u>

It is a polygon which has all the sides of equal length and all the angles of equal measure.

Also the perimeter of the regular polygon is given as the sum of all the sides of the regular polygon.

i.e. the formula for the perimeter is given by:

Perimeter=ns

where n is the number of sides in a given polygon and s is the length of each sides of the polygon.

Here we have a polygon as a pentagon.

i.e.

n=5

and Perimeter=100 inches

i.e.

100=5s\\\\i.e.\\\\s=\dfrac{100}{5}\\\\i.e.\\\\s=20\ inches

Sergio [31]3 years ago
6 0
A regular pentagon is a polygon with five sides of equal length.

The perimeter of a regular polygon is the number of sides (n) multiplied by the length of one side (s).

P=ns
100=5s
s=20

The length of each side is 20 inches

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Jim had some dimes, quarters, and nickels. First, he counted only the dimes and quarters and found that he had 9 coins. Then, he
Olegator [25]

Answer:

$1.90

Step-by-step explanation:

Represent the numbers of nickels, dimes and quarters by n, d and q.

Then: d + q = 9, or q = 9 - d.

Also, n + d = 10

Lastly, n + d + q = 15.

Let's substitute 9 - d for q in the equation immediately above:

n + d + (9 - d) = 15, or n + 9 = 15.  Then n must be 6.

In summary, Jim has 6 nickels, (10 - 6) dimes and (9 - 4) quarters, or:

                                    6 nickels, 4 dimes and 5 quarters

Thus, he has 5($0.05) + 4($0.10) + 5($0.25), or

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7 0
3 years ago
1.) Determine the type of solutions for the function (Picture 1)
NNADVOKAT [17]

Answer:

1) 2 nonreal complex roots

2) 1 Real Solution

3) 16

4) Reflected, narrower by a factor of 2/5, slides right 4 units and slides up 6 (units)

Step-by-step explanation:

1) The graph does not intercept the x-axis, therefore, there are no real solutions at the point y = 0

We get;

y = a·x² + b·x + c

At y = 6, x = -2

Therefore;

6 = a·(-2)² - 2·b + c = 4·a - 2·b + c

6 = 4·a - 2·b + c...(1)

At y = 8, x = 0

8 = a·(0)² + b·0 + c

∴ c = 8...(2)

Similarly, we have;

At y = 8, x = -4

8 = a·(-4)² - 4·b + c = 16·a - 4·b + 8

16·a - 4·b = 0

∴ b = 16·a/4 = 4·a

b = 4·a...(3)

From equation (1), (2) and (3), we have;

6 = 4·a - 2·b + c

∴ 6 = b - 2·b + 8 = -b + 8

6 - 8 = -b

∴ -b = -2

b = 2

b = 4·a

∴ a = b/4 = 2/4 = 1/2

The equation is therefor;

y = (1/2)·x² + 2·x + 8

Solving we get;

x = (-2 ± √(2² - 4 × (1/2) × 8))/(2 × (1/2))

x =( -2 ± √(-12))/1 = -2 ± √(-12)

Therefore, we have;

2 nonreal complex roots

2) Give that the graph of the function touches the x-axis once, we have;

1 Real Solution

3) The given function is f(x) = 2·x² + 8·x + 6

The general form of the quadratic function is f(x) = a·x² + b·x + c

Comparing, we have;

a = 2, b = 8, c = 6

The discriminant of the function, D = b² - 4·a·c, therefore, for the function, we have;

D = 8² - 4 × 2 × 6 = 16

The discriminant of the function, D = 16

4.) The given function is g(x) = (-2/5)·(x - 4)² + 6

The parent function of a quadratic equation is y = x²

A vertical translation is given by the following equation;

y = f(x) + b

A horizontal to the right by 'a' translation is given by an equation of the form; y = f(x - a)

A vertical reflection is given by an equation of the form; y = -f(x) = -x²

A narrowing is given by an equation of the form; y = b·f(x), where b < 1

Therefore, the transformations of g(x) from the parent function are;

g(x) is a reflection of the parent function, with the graph of g(x) being narrower by 2/5 than the graph of the parent function. The graph of g(x) is shifted right by 4 units and is then slides up by 6 units.

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Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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