Let a and b be real numbers where a ≠ b ≠ 0. which of the following functions could represent the graph below?
(x) = x(x – a)2(x – b)4

f(x) = x(x – a)3(x – b)2

f(x) = x4(x – a)(x – b)2

f(x) = x2(x – a)5(x – b)

The correct answer is option (C), that is,

Step-by-step explanation:

From the given graph, we can see that the left most point where graph meets axis is the point where graph touches the x axis. Therefore, the factor corresponding to that point must have an even multiplicity. Therefore, exponent of one of (x-a) or (x-b)  must be even.

Further, at x=0, the graph touches x axis. Therefore, multiplicity of this zero must be even as well. This implies that exponent of x must be even.

Finally, at right most zero, we can see that the graph cross x axis, without staying on x axis. Therefore, multiplicity of that particular zero must be odd and more accurately it should be 1.

Combining all these factors together, we can conclude that one of (x-a) and (x-b) must have an exponent 1 and other must have an even exponent. At the same time x must have an even exponent. All these conditions are satisfied by the function . Here, x has an even exponent (4). (x-a) has exponent 1 and (x-b) has an even exponent (2). Therefore, correct answer is option (c).

true

step-by-step explanation:

∠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.