The solution to the problem is as follows:
We have 2+.8(2) + .8(.8(2)) + .8(.8(.8(2))) + ... =
2( .8^0 + .8^1 + .8^2 + .8^3 + ... ) =
2(.8^n -1) / (.8-1) . As n-->infinity, .8^n-->0 giving us
<span>2(-1)/(-.2) = 2(5) = 10 meters.
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Answer:
The number of different combinations of three students that are possible is 35.
Step-by-step explanation:
Given that three out of seven students in the cafeteria line are chosen to answer a survey question.
The number of different combinations of three students that are possible is given as:
7C3 (read as 7 Combination 3)
xCy (x Combination y) is defines as
x!/(x-y)!y!
Where x! is read as x - factorial or factorial-x, and is defined as
x(x-1)(x-2)(x-3)...2×1.
Now,
7C3 = 7!/(7 - 3)!3!
= 7!/4!3!
= (7×6×5×4×3×2×1)/(4×3×2×1)(3×2×1)
= (7×6×5)/(3×2×1)
= 7×5
= 35
Therefore, the number of different combinations of three students that are possible is 35.
2times1 because 2.75 take the 75 off and 1.25 take the 25 off so you multiply 2 and 1 so 2times1 equals 2
-3x-2.5=y would be an equivalent to that equation
