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Mathematics 7 Online
jas2024:

I forgot what to do ... SS below

jas2024:

1 attachment
xxemilianaxx:

To find the arc length for ex #4 r= 14 yd. 0 = 225 do \[225.(14yrd).\frac{ 2\pi}{ 360 }\]

jas2024:

so 54.95 ?

xxemilianaxx:

Evaluated that problem would be \[\frac{ 35dy \pi }{ 2}\]

jas2024:

i got the same answer...

jas2024:

Well thank you . I think I got the rest.

Florisalreadytaken:

4 is asking you to find the AREA of it whatever that person above did, its wrong

Florisalreadytaken:

it would look like this if you were to solve it right: \[ A=\frac{\theta}{360}\times \pi \times r^2 \] thus, \[ A= \frac{225}{360}\times 3.14 \times 14^2 \] \[ A=\frac{225}{360}\times 3.14 \times 196 \] \[ A=\frac{225}{360}\times 3.14 \times 196 \] \[ A=\frac{225}{360}\times 615.44 \] \[ A \approx 384.7 \ \ yd^2 \]

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