The length of a rectangle is increasing at a rate of 8cm/s8cm/s and its width is increasing at a rate of 5cm/s5cm/s. When the length is 20cm20cm and the width is 25cm,25cm, how fast is the area of the rectangle increasing

Step-by-step explanation:

Given that the length of a rectangle is increasing at a rate of 8cm per s and its width is increasing at a rate of 5cm per s.

Let l be the Length of Rectangle (cm)

Let w  be the Width of Rectangle (cm)

Let A  be the Area of Rectangle ()

Let t  be the Time (s)

From the given we can write

cm per s and  

cm per s

The formula for Area of the rectangle is:

A=lw  square units

Differentiating with respect to t

 ( by using the product rule formula )

when l=20 and w=25

∴  

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}

to find the total number of combinations, multiply the quantity of each item by each other:

2 pairs of shoes x 3 pairs of shorts = 6 different combinations.