Find the value of n such that x2– 3x + n is a perfect square trinomial.

Hello!

First, a word to the wise: Please type x2– 3x + n as x^2– 3x + n. The " ^ " denotes exponentiation, whereas x2 is meaningless.

To make x^2– 3x + n into a perfect square trinomial, take half of the coefficient of x (that is, half of -3) and square the result:

x^2– 3x + (3/2)^2, or x^2– 3x + 9/4. The desired value of n is then 9/4.

i think 1: 2

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)