Total number of ways to make a pair:
The first player can be any one of 7 . For each of those . . .
The opponent can be any one of the remaining 6 .
Total ways to make a pair = 7 x 6 = 42 ways .
BUT ... every pair can be made in two ways ... A vs B or B vs A .
So 42 'ways' make only (42/2) = 21 different pairs.
If every pair plays 2 matches, then (21 x 2) = <em><u>42 total matches</u></em> will be played.
Now, is that an elegant solution or what !
is ≈3.32 and
≈ 6.71. Look at the whole numbers, or integers, between 3 and 6. You will have 4 and 5. SInce we are looking for the sum of the integers, your sum will be 9. Hope this helped.
The second garden plot will require 8√5 feet more fence than the first garden plot.
Further explanation:
In order to find the fence, we have to find the perimeter of both squares
So,
Area of Square 1: A1=180 square feet
Area of Square 2: A2=320 Square feet
Let x be the side of square 1:
Then,
![A_1=x^2\\180=x^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{180}=\sqrt{x^2}\\x=\sqrt{2*2*3*3*5}\\x=\sqrt{2^2*3^2*5}\\x=2*3\sqrt{5}\\x=6\sqrt{5}](https://tex.z-dn.net/?f=A_1%3Dx%5E2%5C%5C180%3Dx%5E2%5C%5CTaking%5C%20square%5C%20root%5C%20on%5C%20both%5C%20sides%5C%5C%5Csqrt%7B180%7D%3D%5Csqrt%7Bx%5E2%7D%5C%5Cx%3D%5Csqrt%7B2%2A2%2A3%2A3%2A5%7D%5C%5Cx%3D%5Csqrt%7B2%5E2%2A3%5E2%2A5%7D%5C%5Cx%3D2%2A3%5Csqrt%7B5%7D%5C%5Cx%3D6%5Csqrt%7B5%7D)
For second square:
Let y be the side of second square
![A_2=y^2\\320=y^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{320}=\sqrt{y^2}\\y=\sqrt{2*2*2*2*2*2*5}\\y=\sqrt{2^2*2^2*2^2*5}\\y=2*2*2\sqrt{5}\\y=8\sqrt{5}](https://tex.z-dn.net/?f=A_2%3Dy%5E2%5C%5C320%3Dy%5E2%5C%5CTaking%5C%20square%5C%20root%5C%20on%5C%20both%5C%20sides%5C%5C%5Csqrt%7B320%7D%3D%5Csqrt%7By%5E2%7D%5C%5Cy%3D%5Csqrt%7B2%2A2%2A2%2A2%2A2%2A2%2A5%7D%5C%5Cy%3D%5Csqrt%7B2%5E2%2A2%5E2%2A2%5E2%2A5%7D%5C%5Cy%3D2%2A2%2A2%5Csqrt%7B5%7D%5C%5Cy%3D8%5Csqrt%7B5%7D)
Perimeter of First Square:
![P_1=4x\\=4(6\sqrt{5})\\=24\sqrt{5}\ feet](https://tex.z-dn.net/?f=P_1%3D4x%5C%5C%3D4%286%5Csqrt%7B5%7D%29%5C%5C%3D24%5Csqrt%7B5%7D%5C%20feet)
Perimeter of Second Square:
![P_2=4y\\=4(8\sqrt{5})\\=32\sqrt{5}\ feet](https://tex.z-dn.net/?f=P_2%3D4y%5C%5C%3D4%288%5Csqrt%7B5%7D%29%5C%5C%3D32%5Csqrt%7B5%7D%5C%20feet)
The smaller perimeter will be subtracted from larger perimeter to find that how much more fence will be needed.
![P_2-P_1=32\sqrt{5}-24\sqrt{5}\\=(32-24)\sqrt{5}\\=8\sqrt{5}\ feet](https://tex.z-dn.net/?f=P_2-P_1%3D32%5Csqrt%7B5%7D-24%5Csqrt%7B5%7D%5C%5C%3D%2832-24%29%5Csqrt%7B5%7D%5C%5C%3D8%5Csqrt%7B5%7D%5C%20feet)
The second garden plot will require 8√5 feet more fence than the first garden plot.
Keywords: Radicals, Operations on Radicals
Learn more about radicals at:
#LearnwithBrainly
Answer:
c
Step-by-step explanation:
F(x)=60+0,75x
Because she charges 60 per session, which will not change. And you don’t know how many miles she has to drive so that will be 0,75x.