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Natalka [10]
3 years ago
8

Find the perimeter of a regular octagon if one side has a measure of 8cm A) 16 cm B) 56cm C) 64cm D) 72cm

Mathematics
2 answers:
zysi [14]3 years ago
8 0
C. 64cm 
Hope this helps! :)
Vesna [10]3 years ago
5 0
C. 64cm. P=8a=8·8=64. I hope this helps you.
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Answer:the answer is Y= -4/3x - 5/2

Step-by-step explanation:

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I’ll give brainliest i’m in middle school<br> help
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The answer would be Y=5x-5
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If a sequence of 8 consecutive odd integers with increasing values has 9 as its seventh term. What is the sum of the terms of th
klemol [59]

Answer:

Sum = 32

Step-by-step explanation:

Given

Sequence = Consecutive Odds

T_7 = 9

Required

The sum of the terms

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S = \{-3,-1,1,3,5,7,9,11\}

Add up all terms in the sequence

Sum = -3-1+1+3+5+7+9+11

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3 years ago
A ribbon is 1.28 meters long. A rope is 2.34 meters longer than the ribbon. How long
satela [25.4K]

Answer:

4.02m

Step-by-step explanation:

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1 . 2 8

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3 years ago
The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width
blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given  perimeter

perimeter of rectangle = 2 ( length +  width )

=> 30.8 km = 2 ( x + y )  

=>x + y = \frac{30.8}{2}

=> y = 15.4 – x             ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,  

=> length^2 + width^2 = diagonal^2

=> x^2 + y^2 = 11^2

=> x^2 + y^2 = 121

On substituting value of y from (1) in above equation we get

=> x^2 + (15.4-x)^2 = 121

=>x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x   = 121

=> 2x^2-30.8x + 237.16 -121  = 0

=> 2x^2-30.8x + 116.16 = 0

Solving above quadratic equation using quadratic formula

General form of quadratic equation is  

ax^2 +bx +c = 0

And quadratic formula for getting roots of quadratic equation is  

x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}

As equation is 2x^2-30.8x + 116.16 = 0, in our case

a = 2 ,  b = -30.8 and c = 116.16

Calculating roots of the equation we get

x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}

x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{(30.8\pm29.33)}4

x=\frac{(30.8+29.33)}{4}

x=\frac{(30.8-29.33)}{4}

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,  

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0
3 years ago
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