The number behind the zero is bigger than 4 so you would round the zero up to a 1.
The answer would be 32.1
Answer:
try with
![77 - 99 + 33 \div 6](https://tex.z-dn.net/?f=77%20-%2099%20%20%2B%2033%20%5Cdiv%206)
to calculate so you can find the answer thanks me later
Step-by-step explanation:
<h2>
<em><u>carry </u></em><em><u>on </u></em><em><u>learning</u></em></h2>
Answer:
Step-by-step explanation:
we have a circle inscribed in a square
Area of the square
Asquare=14*14=196
Acircle=3.14*r^2
r=14/2=7
Acircle=3.14*7^2=153.94
A shaded=196-153.94=42.06
The first area is simply the are of the innermost circle, so we have
![A_1 = \pi r_1^2](https://tex.z-dn.net/?f=%20A_1%20%3D%20%5Cpi%20r_1%5E2%20)
Then, the region inside circle 2 and outside circle 1 is the difference between the areas of these circles:
![A_2 = \pi r_2^2 - \pi r_1^2 = \pi(r_2^2-r_1^2)](https://tex.z-dn.net/?f=%20A_2%20%3D%20%5Cpi%20r_2%5E2%20-%20%5Cpi%20r_1%5E2%20%3D%20%5Cpi%28r_2%5E2-r_1%5E2%29%20)
By the same logic, we have
![A_3 = \pi r_3^2 - \pi r_2^2 = \pi(r_3^2-r_2^2)](https://tex.z-dn.net/?f=%20A_3%20%3D%20%5Cpi%20r_3%5E2%20-%20%5Cpi%20r_2%5E2%20%3D%20%5Cpi%28r_3%5E2-r_2%5E2%29%20)
So, the ratios are
![\dfrac{A_1}{A_2} = \dfrac{\pi r_1^2}{\pi(r_2^2-r_1^2)} = \dfrac{r_1^2}{r_2^2-r_1^2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7BA_1%7D%7BA_2%7D%20%3D%20%5Cdfrac%7B%5Cpi%20r_1%5E2%7D%7B%5Cpi%28r_2%5E2-r_1%5E2%29%7D%20%3D%20%5Cdfrac%7Br_1%5E2%7D%7Br_2%5E2-r_1%5E2%7D%20)
And similarly
![\dfrac{A_2}{A_3} = \dfrac{\pi(r_2^2-r_1^2)}{\pi(r_3^2-r_2^2)} = \dfrac{r_2^2-r_1^2}{r_3^2-r_2^2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7BA_2%7D%7BA_3%7D%20%3D%20%5Cdfrac%7B%5Cpi%28r_2%5E2-r_1%5E2%29%7D%7B%5Cpi%28r_3%5E2-r_2%5E2%29%7D%20%3D%20%5Cdfrac%7Br_2%5E2-r_1%5E2%7D%7Br_3%5E2-r_2%5E2%7D%20)