Are triangles adc and ebc congruent? triangle adc is overlapping triangle ebc so that they both share angle ace, point b is on segment ac, point d is on segment ec, segments ad and eb intersect at point f, segments bc and dc are each marked with one dash, and angles cad and ceb are each marked with an arc yes, by sss yes, by aas yes, by sas not enough information

Yes, by AAS ⇒ 2nd answer

Step-by-step explanation:

* Lets revise the cases of congruence

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 corresponding sides in the 2nd Δ  

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 corresponding

 sides and   including angle in the 2nd Δ  

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ   ≅

 2 corresponding angles and the side whose joining them in the 2nd Δ  

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 corresponding

 angles   and one side in the 2ndΔ

* Lets solve the problem

- In 2 Δs ADC and EBC

∵ ∠ACE is a common angle in the two triangles ⇒ given

∵ ∠CAD ≅ ∠CEB ⇒ because they marked with the same symbol

∵ DC ≅ BC ⇒ because they marked with the same symbol

- Two angles and one side in Δ ADC ≅ two corresponding angles  

  and one side in Δ EBC

∴ Triangles ADC and EBC congruent by AAS

* Yes Triangles ADC and EBC congruent by AAS

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).