What are the vertical and horizontal asymptotes for the function (x)= 3x^2/x^2-4
horizontal asymptote: y = -2, y = 2
vertical asymptote: x= 3
horizontal asymptote: y = 4, y = 1
vertical asymptote: x = 3
horizontal asymptote: y = 3
vertical asymptote: x = 4, x = 1
horizontal asymptote: y = 3
vertical asymptote: x = -2, x = 2

answer: i can’t tell you that

step-by-step explanation:

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc

option: d is the correct answer.

d) none of the above.

step-by-step explanation:

clearly from the figure attached to the question we could see that there is no shape given. hence without any shape given in the figure and after rotation we can't obtained a square pyramid.

since, to obtain a image we need a proper figure of the pre-image so that it can be transformed.

hence, the correct answer is:

d) none of the above.

step-by-step explanation:

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct