The length of a rectangle is 5 centimeters less than six times its width. its area is 14 square centimeters. find the dimensions of the rectangle.

The length is 2.2 centimetres   and the width is 6.5 centimetres

Step-by-step explanation:

Let l be the length of the rectangle and w be the width of the rectangle.

First we will write the statement mathematically

"the length of a rectangle is 5 centimetres less than six times its width" can be written as ;

l = 5 - 6w (1)

"its area is 14 square centimetres" can be written as;

l × w =  14 (2)

substitute equation (1) in equation (2)

(5-6w)w = 14

5w - 6w² = 14

The equation can be re-arrange as;

6w² - 5w -14=0

Using formular method to solve;

] = 6 b = -5   c = -14

x =- b ±√b² - 4ac   /2

x = -6  ±√25 + 336

x = -6± √361

x = -6±19/2

Either x = -6 + 19 or  x = -6 -19

x=  13/2   or   x = -25/2

There is no such thing as negative length, so we will take w = 13/2

w = 13/2

Substitute w = 13/2 in equation (2)

l × w = 14

l×13/2 = 14

13l / 2 = 14

multiply both-side by 2

13l = 28

Divide both-side by 13

l ≈ 2.2

Therefore,

The width = 13/2 = 6.5  and the length is 2.2

why do wages differ across industries

step-by-step explanation :

micro economists are concerned with economics on an industry or individual firm basis.

macro economists are concerned with national and international level

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).