Alandowner wishes to use 3 miles of fencing to enclose an isosceles triangular region of as large an area as possible. what should be the lengths of the sides of the triangle? let x be the length of the base of the triangle. write the area as a function of x. [first write the length of the equal-length sides in terms of the base, x, then write the height of the triangle in terms of the base.]

The lengths of each side = 1 mile

The largest area = 1/4(√3) miles²

Step-by-step explanation:

∵ Th length of the base = x

∵ Perimeter of Δ = 3 miles

∵ The triangle is isosceles

∴ The length of the two equal sides = (3 - x)/2

∵ The height of the Δ is ⊥ to the base and bisect it

∴ h² = [(3 - x)/2]² - (x/2)² = (9 - 6x + x²)/4 - x²/4

∴ h² = 9/4 - 3/2 x + x²/4 - x²/4 = 9/4 - 3/2 x

∴ h = √(9/4 - 3/2 x) = √[(9 - 6x)/4] = 1/2[√(9 - 6x)]

∴ Area of the Δ = (1/2) (x) [1/2√(9 - 6x)]

∴ A = x/4 [√(9 - 6x) ]

∴

∵ The largest area of the given perimeter is the area

   of the equilateral triangle

∴ The length of each side = 3/3 = 1 mile

∴ The largest area = 1/4(√3) miles²

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

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).