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Arte-miy333 [17]
3 years ago
8

Each side of a regular polygon is 5.2 m and its perimeter is 36.4 m. Find the number of sides of the polygon.

Mathematics
1 answer:
fgiga [73]3 years ago
8 0

Answer:

Number of sides of the polygon is 7.

Step-by-step explanation:

Let x be the number of sides of the polygon.

Given ,

Length of each side = 5.2 m

Perimeter = 36.4 m

=> 5.2*x = 36.4

x = 36.4/5.2

x = 7

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Margarita [4]

Answer:

x    -4    -3  -2  -1    0    1   2   3    4

y  -54 -20  -4   0   -2  -4  0   16   50

at 4 it is  maximum .

maximum value=50

hope that helps

7 0
3 years ago
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Three more than twice a number r
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Step-by-step explanation:

probably

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3 years ago
Jada rewrites 5•3^x as 15x. Do you agree with Jada that these are equivalent expressions? Explain your reasoning. Help me please
Veseljchak [2.6K]

Answer:yes

Step-by-step explanation:

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7 0
3 years ago
A farmer has a 100 ft by 200 ft rectangular field that he wants to increase by 15.5% by cultivating a strip of uniform width aro
Advocard [28]

Answer:

(a) The strip should be 5ft wide

(b) The strip around the outside field is 10ft wide.

Step-by-step explanation:

Given:

Length of the rectangular field, L= 200 ft

width of the rectangular field, w = 100 ft

Area of the rectangular field, A = 200ft x 100ft = 20000 ft^2

let the width of the strip = x

The strip around the outside field = 2x

If the field is increased by 15.5%

New area of the field = 1.155 x 20000 = 23,100 ft^2

The increase in area of the field = 3,100 ft

3,100 = New area of field - old area of the field

3100 = (200 + 2x)(100 + 2x) - 20000

3100 = 20000 + 400x 200x + 4x^2 - 20000

3100 = 600x + 4x^2

Divide through by 4

775 = 150x + x^2

x^2 + 150x - 775 = 0

Factorize

(x + 155)(x-5) = 0

x = 5 ft

The strip should be 5ft wide.

The strip around the outside field = 2 x 5 ft = 10 ft

Thus, the strip around the outside field is 10ft wide.

3 0
3 years ago
This problem I can’t seem to figure out
allochka39001 [22]

Answer:

39 touchdowns and 13 fieldgoals

Step-by-step explanation:

Let t= touchdowns

f = fieldgoals

They scored 35 times

t+f = 35

Touchdown is 7 pts and fieldgoal is 3 pt

7t+3f = 193

Multiply the first equation by-7

-7t -7f =-245

Add this to the second equation

-7t -7f =-245

7t+3f = 193

----------------------

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Divide by -4

-4f/-4 = -52/-4

f = 13

Now we can find t

t+f = 35

t+13 = 52

Subtract 13 from each side

t+13-13 =52-13

t =39

6 0
3 years ago
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