Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 1000 cm3, the pressure is 80 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

The volume is decreasing at a rate of 250 cm^3/min

Step-by-step explanation:

From Boyle's law, the relationship between pressure (P) and volume (V) is inverse in which increase in one quantity (P) leads to a corresponding decrease in the other quantity (V).

Pressure at a certain instant (time) = 80 kPa

Rate at which pressure is increasing = 20 kPa/min

Time required for this increase = pressure ÷ rate = 80 ÷ 20 = 4 min

Volume = 1000 cm^3

Rate at which volume is decreasing at the same time pressure is increasing = 1000 ÷ 4 = 250 cm^3/min

c

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}