Given: Lines p and q are parallel and r is a transversal. Prove: ∠2 ≅ ∠7

Parallel lines p and q are cut by transversal r. On line p where it intersects with line r, 4 angles are created. Labeled clockwise, from the uppercase left, the angles are: 1, 2, 4, 3. On line q where it intersects with line r, 4 angles are created. Labeled clockwise, from the uppercase left, the angles are: 5, 6, 8, 7.
A 2-column table with 4 rows. Column 1 is labeled statements with the entries p is parallel to q and r is a transversal, A, B, angle 2 is congruent to angle 7. Column 2 is labeled reasons with the entries given, vertical angles are congruent, correlated angle theorem, transitive property.

Which statements could complete the proof?

The statement A is Angle 2 is congruent to angle 3

The statement is Angle 3 is congruent to angle 7

Step-by-step explanation:

The given information are;

The given parallel lines = p and q

The given transversal = r

The angles from the top left at the intersection of p and r = 1, 2, 4, 3

The angles from the top left at the intersection of q and r = 5, 6, 8, 7

The 2-column table is presented as follows;

Statement                                                          Reasons

p is parallel to q and r is a transversal             Given

A                                                                          Vertical angles are congruent

B                                                                          Correlated angle theorem

Angle 2 is congruent to angle 7                        Transitive property

We have the orientation of the angles given as follows;

1, 2

3, 4

5, 6

7, 8

Therefore, we have;

A = Angle 2 is congruent to angle 3   Vertical angles are congruent

B = Angle 3 is congruent to angle 7     Correlated angle theorem

To give;

Statement                                                          Reasons

p is parallel to q and r is a transversal             Given

Angle 2 is congruent to angle 3                       Vertical angles are congruent

Angle 3 is congruent to angle 7                       Correlated angle theorem

Angle 2 is congruent to angle 7                       Transitive property

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).

step-by-step explanation:

graph using a few selected

x         y

-1       15

0         6

1         2.4

2       0.96