Can you guys help we with this question.

40

Step-by-step explanation:

(2x+1/(2x))^5 *(2x -1/(2x))^5

= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2

= (4x^2-1/4(x)^2)^5

now

x =4x^2. ,a = 1/4(x)^2 ,n =5

we have

general term = Cr *x^r *a^(n-r)

= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)

= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)

= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)

= Cr * 4(2r-n) *x(4r-2n)

now for x^2

4r-2n = 2

4r -10=2

4r =12

r = 3

now for coeff

C(5,3) * 4^(2*3-5)

5!/(3!*(5-3)!) * 4

5*4/(2*1)*4

40

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.