NEED HELP W GEOMETRY In △
ABC, ∡A
is 2 more than twice ∡B
, and ∡C
is 7 less than twice ∡B
. What is the measure of each angle in △
ABC and what is its classifications?

Answers:

Triangle ABC is an acute triangle. It's also a scalene triangle.

Work Shown:

x = measure of angle B

2x+2 = measure of angle A, since we're told its "2 more than twice angle B"

2x-7 = measure of angle C, for similar reasoning

The three angles add to 180. This is true for any triangle

A+B+C = 180

(2x+2)+(x)+(2x-7) = 180

5x-5 = 180

5x = 180+5

5x = 185

x = 185/5

x = 37

This means angle B is 37 degrees. The other two angles are

A = 2x+2 = 2*37+2 = 74+2 = 76

C = 2x-7 = 2*37-7 = 74-7 = 67

In summary, we have these three interior angles

As a way to check, note how A+B+C = 76+37+67 = 180. This confirms the answer. All three angles mentioned are smaller than 90 degrees, so this triangle is acute.

Furthermore, because all three angles are different measures, this means all three sides are different lengths. This leads to triangle ABC being scalene.

what "following statement"?

step-by-step explanation:

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)

=====================================

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

=====================================

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).