Find the values of the six trigonometric functions for angle ѳ, when pq=48 and qr=64

Here's what I get    

Step-by-step explanation:

∆PQR is a right triangle.

PR² = PQ² + QR² = 48² + 64² = 2304 + 4096 = 6400

PR = √6400 = 80  

PQ:QR:PR = 48:64:80 = 3:4:5

We can consider ∆PQR as a 3:4:5 triangle.

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

step-by-step explanation: