Find all values of k such that the graph of the inequality (x-2)(x-5)(x-k) >= 0 consists of a single interval on the number line.

(x - 2)(x - 5)(x - k) >= 0

Let's set each term equal to 0

x - 2= 0

Add 2 to both sides

x = 2.

x - 5 = 0

Add 5 to both sides

x = 5

x - k = 0

Add k to both sides

x = k

Since x = 2 and 5, k = 2, 5

Let's plug our numbers into the inequality.

(2 - 2)(2 - 5)(2 - 2) >= 0

(0)(-3)(0) >= 0

0>=0

(5 - 2)(5 - 5)(5 - 5) >= 0

(3)(0)(0) >= 0

0 >= 0

This proves that 2 and 5 are equal to k

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

4.

step-by-step explanation:

it have the same area.brainliest