Ahummingbird's nest is 6 meters high in a tree and a flower is on the ground 4 meters away from the base of the tree. how far will the hummingbird need to fly to get from its nest to the flower? if necessary round to the nearest tenth.

7.2 metres

Step-by-step explanation:

For this problem, it's best to draw a diagram to help you visualize. See below for a diagram. I used H for hummingbird; T for tree, F for flower. We are looking for "t", the blue side.

Label sides using the opposite angle's letter, but use its lowercase.

Since this is a right triangle and we know two of the three sides, use the Pythagorean Theorem. The general formula is:

a² + b² = c²

"c" means the hypotenuse, or the longest side. "a" and "b" are the other non-hypotenuse sides, called 'legs'.

In this problem, our legs are f = 6 and h = 4. The hypotenuse is "t".

Write the formula and solve for "t".

h² + f² = t²                Substitute the values f = 6 and h = 4

4² + 6² = t²              Square the numbers

16 + 36 = t²             Add

52 = t²                Put the variable on the left side as standard format

t = √52              Square root both sides

t ≈ 7.2                 Rounded answer

Remember to add in the units.

t = 7.2 m

Therefore, the hummingbird will need to fly 7.2 metres to get from its nest to the flower.

number one is the correct answer

step-by-step explanation: dogs and cats are mentioned, no other animals

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).