What is k-4>= 15\lor k-16<1 ?

The solution to k-4>= 15\lor k-16 = 19

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k-4>= 15\lor k-16<1

k - 4 \geq 15 or k - 16 < 1

k < 17 or k \geq 19

Interval Notation

(- \infty, 17) \cup [19, \infty)

Solution steps

k - 4 \geq 15 or k - 16 < 1

k - 4 \geq 15 : k \geq 19

k - 16 < 1 : k < 17

Combine the intervals

k \geq 19 or k < 17

Merge Overlapping Intervals

k < 17 or k \geq 19

2 < ∣ x + 3 ∣ \leq 6

(8 c^{2} d) (10 c^{3} d^{5})

\frac{( 5 x - 6 )}{13} = 0

x^{2} - 7 x + 12.25

What is k-4>= 15\lor k-16<1 ?
The solution to k-4>= 15\lor k-16<1 is k<17\lor k>= 19