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noname [10]
3 years ago
11

What is the arc length if θ = 7 pi over 4 and the radius is 5?

Mathematics
2 answers:
PolarNik [594]3 years ago
4 0

The formula for arc length is:

s= r∅

Given that ∅= 7π/4

and r= 5

Therefore, arc length s= 7π/4 *5 = 35π/4

jarptica [38.1K]3 years ago
4 0

Answer:  The required arc length will be 27.5 units.

Step-by-step explanation:  We are given to find the arc length with the following information :

\theta=\dfrac{7\pi}{4},~~~\textup{radius},~r=5.

We know that the length of an arc with angle subtended at the center α and radius of the circle r units is given by

\ell=r\alpha.

Therefore, the required arc length will be

\ell=r\theta=5\times\dfrac{7\pi}{4}=\dfrac{35}{4}\times \dfrac{22}{7}=\dfrac{5\times 11}{2}=\dfrac{55}{2}=27.5.

Thus, the required arc length will be 27.5 units.

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vlada-n [284]

1.

(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1

2.

(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10

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