Answer:
I=PRT is the formula
Step-by-step explanation:
$3600 is the Principal. 9% is the interest rate. 25 years is the Time.
This is all I can help you with. Hope you get it write and good luck.
Answer:
x = 7 1/3
Step-by-step explanation:
First, add 1/2 and 6
1/2 + 12/2 = 13/2 = 6 1/2
Our equation now:
13/2 over 3 - x - 3 over 2 = 0
Next, divide 13/2 by 3
13/2 divided by 3 = 13/6
Our equation now:
13/6 over 3 - (x - 3) over 2 = 0
Now, find common denominators and simplify
13/6 - ((x - 3) times 3) over 6 = 0
22 - 3x over 6 = 0
Multiply by six on both sides
22 - 3x = 0
Subtract 22 from both sides
-3x = -22
Divide by -1
3x = 22
Divide both sides of the equation by 3
x = 22/3
x = 7 1/3
Hope this helps!
<u>Answer-</u>
<em>For </em><em>x = 2.24</em><em>, greatest possible amount of light will be admitted.</em>
<u>Solution-</u>
According to the question, the perimeter of the window is 16 ft
From the attachment,
Perimeter of the window = Perimeter of the rectangle-Width of the rectangle + Curved perimeter of the semi-circle
So,
![\Rightarrow (2x+y+y+2x)-(2x)+(\pi x)=16\\\\\Rightarrow 2x+2y+\pi x=16\\\\\Rightarrow 2y=16-2x-\pi x\\\\\Rightarrow y=\dfrac{16-2x-\pi x}{2}\\\\\Rightarrow y=8-x-\dfrac{\pi x}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20%282x%2By%2By%2B2x%29-%282x%29%2B%28%5Cpi%20x%29%3D16%5C%5C%5C%5C%5CRightarrow%202x%2B2y%2B%5Cpi%20x%3D16%5C%5C%5C%5C%5CRightarrow%202y%3D16-2x-%5Cpi%20x%5C%5C%5C%5C%5CRightarrow%20y%3D%5Cdfrac%7B16-2x-%5Cpi%20x%7D%7B2%7D%5C%5C%5C%5C%5CRightarrow%20y%3D8-x-%5Cdfrac%7B%5Cpi%20x%7D%7B2%7D)
As we have to find the value of x so that the greatest possible amount of light is admitted, so we have to calculate the maximum area for which the perimeter is 16 ft.
![\text{Area of the window}=\text{Area of the rectangle + Area of the semi-circle}](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20the%20window%7D%3D%5Ctext%7BArea%20of%20the%20rectangle%20%2B%20Area%20of%20the%20semi-circle%7D)
So,
![A=(2x\times y)+(\dfrac{\pi x^2}{2})](https://tex.z-dn.net/?f=A%3D%282x%5Ctimes%20y%29%2B%28%5Cdfrac%7B%5Cpi%20x%5E2%7D%7B2%7D%29)
Putting the value of y in terms of x,
![\Rightarrow A=2x(8-x-\dfrac{\pi x}{2})+\dfrac{\pi x^2}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20A%3D2x%288-x-%5Cdfrac%7B%5Cpi%20x%7D%7B2%7D%29%2B%5Cdfrac%7B%5Cpi%20x%5E2%7D%7B2%7D)
![\Rightarrow A=16x-2x^2-\pi x^2+\dfrac{\pi x^2}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20A%3D16x-2x%5E2-%5Cpi%20x%5E2%2B%5Cdfrac%7B%5Cpi%20x%5E2%7D%7B2%7D)
![\Rightarrow A=16x-2x^2-\dfrac{\pi x^2}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20A%3D16x-2x%5E2-%5Cdfrac%7B%5Cpi%20x%5E2%7D%7B2%7D)
![\Rightarrow A'=16-4x-\pi x](https://tex.z-dn.net/?f=%5CRightarrow%20A%27%3D16-4x-%5Cpi%20x)
Now, equating the first derivative of A to 0,
![\Rightarrow A'=0](https://tex.z-dn.net/?f=%5CRightarrow%20A%27%3D0)
![\Rightarrow 16-4x-\pi x=0](https://tex.z-dn.net/?f=%5CRightarrow%2016-4x-%5Cpi%20x%3D0)
![\Rightarrow 16-x(4+\pi)=0](https://tex.z-dn.net/?f=%5CRightarrow%2016-x%284%2B%5Cpi%29%3D0)
![\Rightarrow x(4+\pi)=16](https://tex.z-dn.net/?f=%5CRightarrow%20x%284%2B%5Cpi%29%3D16)
![\Rightarrow x=\dfrac{16}{4+\pi}](https://tex.z-dn.net/?f=%5CRightarrow%20x%3D%5Cdfrac%7B16%7D%7B4%2B%5Cpi%7D)
![\Rightarrow x\approx 2.24](https://tex.z-dn.net/?f=%5CRightarrow%20x%5Capprox%202.24)
Therefore, for x= 2.24, greatest possible amount of light will be admitted.