Find two vectors v1 and v2 whose sum is 〈−3,4〉〈−3,4〉, where v1 is parallel to 〈3,−3〉〈3,−3〉 while v2 is perpendicular to 〈3,−3〉〈3,−3〉.

∵ Vector is parallel to 〈3, -3〉,

⇒ = x〈3, -3〉 = 〈3x, -3x〉,

Where, x is any scalar,

According to the question,

= 〈-3,4〉

⇒ 〈3x, -3x〉+ = 〈-3,4〉

⇒ = 〈-3,4〉 - 〈3x,-3x〉

Now, is perpendicular to 〈3, -3〉,

⇒ .〈3, -3〉 = 0

⇒ ( 〈3,4〉 - 〈3x, -3x〉 ). 〈3, -3〉 = 0

⇒ 〈-3-3x,4+3x〉 . 〈3, -3〉 = 0

⇒ (-3-3x)(3) + (4+3x)(-3) = 0

⇒ - 9 - 9x - 12 - 9x = 0

⇒ -21 - 18x = 0

⇒ x =

Hence, required vectors are,

= 〈-7/2, 7/2〉

=〈3,4〉 - 〈-7/2, 7/2〉 = 〈13/2, 1/2〉

hey seth! : )

i’m not 100% sure, but i think it’s b

hope this ya : p

miss u!

- nicole

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).