In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of ∠K, to the nearest 10th of a degree.

Given:

In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°.

To find:

The value of  ∠K to the nearest 10th of a degree.

Solution:

According to the Law of Sine:

We have, In ΔJKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°.

Using the Law of Sine, we get

Therefore, the measure of angle K is 74.0 degrees.

  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).

step-by-step explanation: