I dare you to solve this!Foil method

a^(2m) + 2 a^m b^ (n-1) + b^ ( 2n -2)

Step-by-step explanation:

( a^m + b^(n-1) )( a^m + b^(n-1) )

FOIL

first: a^m a^m = a^ 2m

outer: a^m b^ n-1

inner a^m b^ n-1

last : b^ n-1  b^ n-1 = b^ ( 2n -2)

Add together

a^(2m) + 2 a^m b^ (n-1) + b^ ( 2n -2)

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is 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.