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bagirrra123 [75]
2 years ago
5

Each side of a hexagon is 10 inches longer than the previous side. what is the length of the shortest side of this hexagon if it

s perimeter is 401 inches?
Mathematics
1 answer:
MrRa [10]2 years ago
3 0

The length of the shortest side of the hexagon is; 41.833 inches

<h3>How to find the perimeter of a Polygon?</h3>

Let the length of the shortest side of the hexagon be x. Now, a hexagon has six sides and if the next side is 10 inches longer than the previous side, then the length of the six sides are;

x, x + 10, x + 20, x + 30, x + 40, x + 50

Perimeter is given as 401 inches. Thus;

x + x + 10 + x + 20 + x + 30 + x + 40 + x + 50 = 401

6x + 150 = 401

6x = 401 - 150

6x = 251

x = 251/6

x = 41.833 inches

Read more about Polygon Perimeter at; brainly.com/question/14490532

#SPJ1

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3 years ago
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A department​ store, on​ average, has daily sales of ​$28 comma 286.28. The standard deviation of sales is ​$1 comma 500. On​ Tu
wolverine [178]

Answer:

z-score = 4.43,

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Step-by-step explanation:

The provided information is:

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Then, x = $34,924.62

The z-score is define as:

z=\dfrac{x-\mu}{\sigma}\\\\=\dfrac{34924.62-28286.28}{1500}\\\\=4.43      

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3 0
3 years ago
Given: PSTK is a trapezoid, m∠P=90° SK =13, PK = 12, ST=8 Find: The area of PSTK.
guajiro [1.7K]
<h3>Given</h3>

trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8

<h3>Find</h3>

the area of PSTK

<h3>Solution</h3>

It helps to draw a diagram.

∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as

... KS² = PS² + KP²

... 13² = PS² + 12²

... PS = √(169 -144) = 5

This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is

... A = (1/2)(b1 +b2)h

... A = (1/2)(12 +8)·5

... A = 50

The area of trapezoid PSTK is 50 square units.

5 0
3 years ago
Jim has received scores of 72 72 and 77 77 on his first two 100 point tests. what score must he get on his third 100 point test
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77+72=149 
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Arte-miy333 [17]
Octagon 
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-MalStar
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4 years ago
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