Answer:
$13500
Step-by-step explanation:
Given
Each tile cover one ninth of a square yard and each tile costs $3
First, area of the Swimming Pool(from the diagram)
=Area of Left rectangle(length=20,breadth=10)+ Area of Square in middle(side=10)+Area of right rectangle(length=20,breadth=10)
![(\text{area of rectangle}= length\times breadth \\\text{ area of square}= side \times side)](https://tex.z-dn.net/?f=%28%5Ctext%7Barea%20of%20rectangle%7D%3D%20length%5Ctimes%20breadth%20%5C%5C%5Ctext%7B%20area%20of%20square%7D%3D%20side%20%5Ctimes%20side%29)
![= 20\times 10+ 10\times 10+ 20 \times 10\\= 200+100+200\\=500yards](https://tex.z-dn.net/?f=%3D%2020%5Ctimes%2010%2B%2010%5Ctimes%2010%2B%2020%20%5Ctimes%2010%5C%5C%3D%20200%2B100%2B200%5C%5C%3D500yards)
We have to calculate how many tiles make up 500 yards
As given each tile is one ninth of a yard
⇒Number of tiles that make up 500 yards, n
![=\frac{500}{\frac{1}{9} } =\frac{500\times9}{1} =4500](https://tex.z-dn.net/?f=%3D%5Cfrac%7B500%7D%7B%5Cfrac%7B1%7D%7B9%7D%20%7D%20%3D%5Cfrac%7B500%5Ctimes9%7D%7B1%7D%20%3D4500)
⇒ As each tile costs $3
Therefore total cost to tile the floor
![= \text{total number of tiles}\times 3\\=n\times 3\\=4500 \times 3\\=13500](https://tex.z-dn.net/?f=%3D%20%5Ctext%7Btotal%20number%20of%20tiles%7D%5Ctimes%203%5C%5C%3Dn%5Ctimes%203%5C%5C%3D4500%20%5Ctimes%203%5C%5C%3D13500)
Therefore total cost of tiles=$13500
Given
![x+1 = \sqrt{7x+15}](https://tex.z-dn.net/?f=x%2B1%20%3D%20%5Csqrt%7B7x%2B15%7D)
We have to set the restraint
![x+1\geq 0 \iff x \geq -1](https://tex.z-dn.net/?f=x%2B1%5Cgeq%200%20%5Ciff%20x%20%5Cgeq%20-1)
because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:
![(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0](https://tex.z-dn.net/?f=%28x%2B1%29%5E2%3D7x%2B15%20%5Ciff%20x%5E2%2B2x%2B1%3D7x%2B15%20%5Ciff%20x%5E2-5x-14%20%3D%200)
The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.
Similarly, we have
![x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}](https://tex.z-dn.net/?f=x-3%20%3D%20%5Csqrt%7Bx-1%7D%2B4%20%5Ciff%20x-7%3D%5Csqrt%7Bx-1%7D)
So, we have to impose
![x-7\geq 0 \iff x \geq 7](https://tex.z-dn.net/?f=x-7%5Cgeq%200%20%5Ciff%20x%20%5Cgeq%207)
Squaring both sides, we have
![(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0](https://tex.z-dn.net/?f=%28x-7%29%5E2%3Dx-1%20%5Ciff%20x%5E2-14x%2B49%3Dx-1%20%5Ciff%20x%5E2-15x%2B50%20%3D%200)
The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.
Answer:
53 and 2/3 calories per ounce
Step-by-step explanation:
<u>p</u><u>r</u><u>o</u><u>p</u><u>o</u><u>r</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u> 480/9 = ?/1
480 divided by 9 = 53 2/3