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yan [13]
3 years ago
8

Calculate the perimeter of this sector. ​

Mathematics
1 answer:
Maru [420]3 years ago
3 0

The two straight edges of the shape correspond to the original circle's radius, 5.5 cm. We're told this sector has area 30.25 cm^2, which we will use to determine the measure of the central angle subtended by the arc (the remaining edge of the shape).

The complete circle has an area of \pi(5.5\,\rm cm)^2, which corresponds to a "central angle" of 2\pi\,\rm rad. Then if \theta is the central angle of this sector, we have

\dfrac{\pi(5.5\,\rm cm)^2}{2\pi\,\rm rad}=\dfrac{30.25\,\mathrm{cm}^2}\theta\implies\theta=2\,\mathrm{rad}

The complete circle has a circumference of 2\pi(5.5\,\rm cm). Then if \ell is the length of the sector's arc, we get

\dfrac{2\pi(5.5\,\rm cm)}{2\pi\,\rm rad}=\dfrac{\ell}{2\,\rm rad}\implies\ell=11\,\rm cm

So the sector has a total perimeter of

5.5 cm + 5.5 cm + 11 cm = 22 cm

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laila [671]

Answer: 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

Step-by-step explanation:

<em>ok</em><em>,</em><em> so just let me</em><em>,</em><em> uh</em><em>,</em><em> make this a little easier on my eyes lol:</em>

-10m^4n^3+8m^2n^6-6m^2n^6 ----> (-10m^4)(n^3) + (8m^2)(n^6) - (6m^2)(n^6)

<em>I just put in parentheses. nothing changed. just parentheses. </em>

Since the leading coefficients, -10 , 8 , and 6 are multiples of two,

we can take out 2 from each... [-10/2 = 5 , 8/2 = 4 , 6/2 = 3 ]

= 2 x [ (-5m^4)(n^3) + (4m^2)(n^6) - (3m^2)(n^6) ]

Looking at it now...

<em>we can take out m^2 from each... </em>m^4 , m^2 , and m^2

m^2 [m^2 , 1 , 1]

since    m^4 /m^2 = m^2 , m^2 /m^2 = 1 , m^2 /m^2 = 1

= 2(m^2) x [ (-5m^2)(n^3) + (4x1)(n^6) - (3x1)(n^6) ]

= 2(m^2) x [ (-5m^2)(n^3) + (4)(n^6) - (3)(n^6) ]

Finally, one more step...

<em>we can take out n^3 from each term.... </em>n^3 , n^6 , and n^6

n^3 [1 , n^2 , n^2] since n^3 /n^3 = 1 , n^6 /n^3 = n^2 , n^6 /n^3 = n^2

= 2(m^2)(n^3) [ (-5m^2)(1) + (4)(n^2) - (3)(n^2) ]

= 2m^2n^3 x [ (-5m^2) + (4n^2) - (3n^2) ]

= 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

ANSWER: 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

<em>keep this parentheses. seriously. </em>

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Hello from MrBillDoesMath!

Answer:

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