The sum of three positive numbers is 1. the difference between the first and second numbers is equal to the third number, while their sum is five times as large as the third number. what is the smallest of these numbers? pls answer quickly, 99 pts

  1/6

Step-by-step explanation:

The three numbers, a, b, c, satisfy the equations ...

Subtracting the last equation from the first gives ...

  (a +b +c) -(a +b -5c) = (1) -(0)

  6c = 1

  c = 1/6

Now, we know the sum of the first two numbers is 5/6 (five times the last), and the difference of the first two numbers is 1/6 (equal to the last). Then the least of the first two numbers is ...

  (5/6 -1/6)/2 = 1/3

and the third number, 1/6, is shown to be the smallest.

The smallest of these numbers is 1/6.

In order, the numbers are 1/2, 1/3, 1/6.

Comment on finding the second-smallest number

For ...

We can find b by subtracting the second equation from the first:

  (a +b) -(a -b) = (5/6) -(1/6)

  2b = 4/6 = 2/3 . . . . . simplify

  b = 1/3 . . . . . . . . . . . . divide by 2

Please note that half the difference of the sum and difference is the generic solution to finding the smallest in a "sum and difference" problem.

i’m sorry but i don’t know sorry

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convert the exponential equation to a logarithmic equation using the logarithm base (6) of the right side (36) equals the exponent (2).

ℓσg6(36)=2

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