Let's start with the area of the square
![area \: square = s \times s = 8 \times 8 = 64](https://tex.z-dn.net/?f=area%20%5C%3A%20square%20%3D%20s%20%5Ctimes%20s%20%3D%208%20%5Ctimes%208%20%3D%2064)
now let's subtract the are of the two half circles.
two half circles are the same as one circle, and we know that the diameter of the circle is 8 (same as the side of a square) so it's radius is 8/2= 4 inches
![area \: circle = \pi {r}^{2} = \pi \times {4}^{2} = 16\pi](https://tex.z-dn.net/?f=area%20%5C%3A%20circle%20%3D%20%5Cpi%20%7Br%7D%5E%7B2%7D%20%20%3D%20%5Cpi%20%5Ctimes%20%20%7B4%7D%5E%7B2%7D%20%20%3D%2016%5Cpi)
now we just subtract and our answer is
The answer is 10 :) ........
Answer:
The average acceleration of the cyclist was 0.6 m/s².
Step-by-step explanation:
Acceleration:
The rate change of velocity per unit time is call the acceleration of the object.
![a=\frac{v-u}t](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7Dt)
u= initial velocity
v= final velocity
t= time taken change of velocity.
During the final 5 seconds of a race cyclist increased her velocity from 4m/s to 7 m/s
Here v= 7 m/s and u=4 m/s t=5 seconds
![\therefore a=\frac{7\ m/s-4 \ m/s}{5s}](https://tex.z-dn.net/?f=%5Ctherefore%20a%3D%5Cfrac%7B7%5C%20m%2Fs-4%20%5C%20m%2Fs%7D%7B5s%7D)
![=\frac{3}{5} \ m/s^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B3%7D%7B5%7D%20%5C%20m%2Fs%5E2)
=0.6 m/s²
The average acceleration of the cyclist was 0.6 m/s².
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