The area of a rectangle is 5/6 square foot. The width is 2/3 foot. How many times greater is the length than the width? Write your answer in simplest form.

15/8

Step-by-step explanation:

The area of a rectangle is given as 5/6 square foot.

The width is given as 2/3 foot

But area if a Rectangle = Lenght × width

A = Lenght x width

5/6= Lenght × 2/3

If we make Lenght subject of the formula we have

lenght= (5/6) / (2/3)

length = (5×3)/(6×2)

Lenght= 5/4

To calculate the number of times greater is the length than the width we take the ratio of Lenght and width

Lenght / width = (5/4) / (2/3)

= (5×3)/(4×2)

= 15/8

Hence the number of times greater is the length than the width is 15/8

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}

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb