The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, … starts with two 1s, and each term afterward is the sum of its two predecessors. Which one of the ten digits is the last to appear in the ones position of a number in the Fibonacci sequence?

6

Step-by-step explanation:

Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $:

$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,$

The last digit to appear in the units position of a number in the Fibonacci sequence is $6.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3