Aglass pitcher is filled with water. 1/8 of the water is poured equally into 2 glasses.
a. what fraction of the water is in each glass?
b. if each glass has 3 fluid ounces of water in it, how many fluid ounces of water were in the full pitcher?
c. if 1/4 of the remaining water is poured out of the pitcher to water a plant, how many cups of water are left in the pitcher?

Step-by-step explanation:

Part a)

Let the volume of water in glass pitcher be 'V'

Now it is given that 1/8 of water is poured in 2 glasses

Thus the volume of water in each glass is

Part b)

Now it is given that volume of water in each glass is 3 fluid ounces thus to obtain the original volume we euate the volume obtained in part a with 3. Thus

Part c)

After pouring out 1/8 water from the orifinal glass the volume of watre that remains in the pitcher euals

Now since 1/4 of the above volume is poured out which euals

Thus the volume of water that remains euals

Applying the value of V we get

ounces

answer: the correct one is a

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}