Amower uses 5/16 gallon of gas to mow the back yard. the side yard is only 2/3 the size of the backyard. how much gas will be needed to mow the side yard?

(5/24) gallons of gas to mow the side yard

Step-by-step explanation:

We can easily solve this proble by applying a rule of three

5/16 gallons of gas    for 1 backyard

x  gallons of gas  2/3 backyard

x = ((2/3)/1)*(5/16) gallons of gas

x = ((2/3))*(5/16) gallons of gas

x = ((10/48) gallons of gas

x = (5/24) gallons of gas

the answer is d

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}