Determine the values of the unlabeled parts of the triangle. OA) ZQ = 63.4°, PQ = 1.84, QO = 4.47

OB) ZQ = 26.6°, PQ = 2, QO = 5.12

OC) ZQ = 63.4º, PQ = 2, QO = 4.47

D)

ZQ = 63.4°, PQ = 1.84, QO = 5.12

∠Q = 63.4º, PQ = 2, QO = 4.47

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.

Triangle POQ is a right angled triangle because ∠P = 90°. Also, ∠O = 26.6°.

Therefore in ΔPOQ: ∠P + ∠O + ∠Q = 180° (sum of angles in a triangle).

90 + 26.6 + ∠Q = 180

∠Q + 116.6 = 180

∠Q = 180 - 116

∠Q = 63.4°

We can find PQ and QO using sine rule. Sine rule for ΔPOQ is given as:

  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).

answ5

step-by-step explanation:

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