Answer:
change below :)
Step-by-step explanation:
You would start out at in the middle of a number line (horizontal). After the first play, you would move to the left on the line 5 spots (to simulate 5 yards lost) and then make a dot to show your new position. You would do the same thing for the next two plays, moving 5 spots to the left each play and creating a new starting point each time. The total spots moved after the 3 plays is 15 spots to the left.
Answer:
bottom side (a) = 3.36 ft
lateral side (b) = 4.68 ft
Step-by-step explanation:
We have to maximize the area of the window, subject to a constraint in the perimeter of the window.
If we defined a as the bottom side, and b as the lateral side, we have the area defined as:
![A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}](https://tex.z-dn.net/?f=A%3DA_r%2BA_c%2F2%3Da%5Ccdot%20b%2B%5Cdfrac%7B%5Cpi%20r%5E2%7D%7B2%7D%3Dab%2B%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cleft%20%28%5Cdfrac%7Ba%7D%7B2%7D%5Cright%29%5E2%3Dab%2B%5Cdfrac%7B%5Cpi%20a%5E2%7D%7B8%7D)
The restriction is that the perimeter have to be 12 ft at most:
![P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12](https://tex.z-dn.net/?f=P%3D%28a%2B2b%29%2B%5Cdfrac%7B%5Cpi%20a%7D%7B2%7D%3D2b%2Ba%2B%28%5Cdfrac%7B%5Cpi%7D%7B2%7D%29%20a%3D2b%2B%281%2B%5Cdfrac%7B%5Cpi%7D%7B2%7D%29a%3D12)
We can express b in function of a as:
![2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a](https://tex.z-dn.net/?f=2b%2B%281%2B%5Cdfrac%7B%5Cpi%7D%7B2%7D%29a%3D12%5C%5C%5C%5C%5C%5C2b%3D12-%281%2B%5Cdfrac%7B%5Cpi%7D%7B2%7D%29a%5C%5C%5C%5C%5C%5Cb%3D6-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29a)
Then, the area become:
![A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2](https://tex.z-dn.net/?f=A%3Dab%2B%5Cdfrac%7B%5Cpi%20a%5E2%7D%7B8%7D%3Da%286-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29a%29%2B%5Cdfrac%7B%5Cpi%20a%5E2%7D%7B8%7D%5C%5C%5C%5C%5C%5CA%3D6a-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29a%5E2%2B%5Cdfrac%7B%5Cpi%20a%5E2%7D%7B8%7D%5C%5C%5C%5C%5C%5CA%3D6a-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D-%5Cdfrac%7B%5Cpi%7D%7B8%7D%5Cright%29a%5E2%5C%5C%5C%5C%5C%5CA%3D6a-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B8%7D%5Cright%29a%5E2)
To maximize the area, we derive and equal to zero:
![\dfrac{dA}{da}=6-2\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right )a=0\\\\\\6-(1-\pi/4)a=0\\\\a=\dfrac{6}{(1+\pi/4)}\approx6/1.78\approx 3.36](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bda%7D%3D6-2%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B8%7D%5Cright%20%29a%3D0%5C%5C%5C%5C%5C%5C6-%281-%5Cpi%2F4%29a%3D0%5C%5C%5C%5Ca%3D%5Cdfrac%7B6%7D%7B%281%2B%5Cpi%2F4%29%7D%5Capprox6%2F1.78%5Capprox%203.36)
Then, b is:
![b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68](https://tex.z-dn.net/?f=b%3D6-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29a%5C%5C%5C%5C%5C%5Cb%3D6-0.393%2A3.36%3D6-1.32%5C%5C%5C%5Cb%3D4.68)
Answer:
Multiply the number by the percent (e.g. 87 * 68 = 5916)
Divide the answer by 100 (Move decimal point two places to the left) (e.g. 5916/100 = 59.16)
Round to the desired precision (e.g. 59.16 rounded to the nearest whole number = 59)
Step-by-step explanation:
lets say for example 234
234/2 = 117 and the percentage were looking for is ummmm 25% well 25% is 1/4th of 234 making it 1/2 of 117 25% of 234 is 58.8 lets do a harder one what is 47% of 234 Divide n by 100.
Multiply the result by x. making it 109.98