Triangle H N K is shown. Angle H N K is 90 degrees. The length of hypotenuse H K is n, the length of H N is 12, and the length of N K is 6. Law of cosines: a2 = b2 + c2 – 2bccos(A) What is the value of n to the nearest whole number? 10 13 18 21

13

Step-by-step explanation:

From the question, we are given a triangle HNK with an angle of 90°

The length of hypotenuse H K is n,

the length of HN is 12

the length of N K is 6.

From the above values, obtained in the question, we can see that this is a right angled triangle.

We are asked to find the length of the hypotenuse.

We can use Pythagoras Theorem of solve for this.

c² = a² + b²

where c = HK = n

a = NK = 6

b = HN = 12

c² = 6² + 12²

c² = 36 + 144

c² = 180

c = √180

c = 13.416407865

Approximately to the nearest whole number = 13

Therefore the value of HK = n = 13

We can also use Law of Cosines as given in the question to solve for this.

a² = b² + c² - 2ac × Cos A

where c = HK = n

a = NK = 6

b = HN = 12

Hence

c² = a² + b² - 2ab × Cos C

c = √ (a² + b² - 2ab × Cos C)

Where C = 90

c = √ 6² + 12² - 2 × 6 × 12 × Cos 90

c = 13.42

Approximately to the nearest whole number ≈ 13

Therefore the value of HK = n = 13

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5