Given m<12= 121 and m<6= 75, find the measure of each missing angle.

Answer/Step-by-step explanation:

Given:

m<12 = 121°

m<6 = 75°

a. m<1 = m<6 (vertical angles)

m<1 = 75° (substitution)

b. m<12 = m<1 + m2 (alternate exterior angles)

121° = 75° + m<2 (substitution)

121° - 75° = m<2 (subtraction property of equality)

46° = m<2

m<2 = 46°

c. m<1 + m<2 + m<3 = 180° (angles on a straight line)

75° + 46° + m<3 = 180° (substitution)

121° + m<3 = 180°

m<3 = 180° - 121° (subtraction property of equality)

m<3 = 59°

d. m<4 = m<3 (vertical angles)

m<4 = 59° (substitution)

e. m<5 + m<4 + m<6 = 180° (angles on a straight line)

m<5 + 59° + 75° = 180° (substitution)

m<5 + 134° = 180°

m<5 = 180° - 134° (Subtraction property of equality)

m<5 = 46°

f. m<7 = m<12 (vertical angles)

m<7 = 121° (substitution)

g. m<8 = m<4 (vertical angles)

m<8 = 59° (substitution)

h. m<9 = m<6 (Alternate Interior Angles)

m<9 = 75° (substitution)

i. m<10 + m<9 = 180° (Linear Pair)

m<10 + 75° = 180° (substitution)

m<10 = 180° - 75° (Subtraction property of equality)

m<10 = 105°

j. m<11 = m<8 (vertical angles)

m<11 = 59° (substitution)

k. m<13 = m<10 (vertical angles)

m<13 = 105° (substitution)

l. m<14 = m<9 (vertical angles)

m<14 = 75° (substitution)

i leave the calculator work to you.

7. diameter 4 miles, circumference 4π miles

8. radius 0.2 miles, circumference 0.4π miles

[this is a bad problem because the actual radius of a car on the track is less than 0.2 miles so the answer is less than 0.4π miles]

9. radius 6 feet, circumference 12π feet

10. diameter 14 cm, circumference 14π cm

11. diameter 4.8 feet, circumference 4.8π feet

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).