$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

8 0

2 years ago

NEED HELP ASAP<br> NO LINKS AND NO JUST SAYING HI, PLEASE ANSWER LEGITIMATELY! THANK YOU!!!! <3

Mekhanik [1.2K]

Answer:

x=42.5

Step-by-step explanation:

First, start by setting up the equation

these are both same side interior angles, so they both add up to 180

(3x+10) + x = 180

4x+10 =180

subtract both sides by 10

-10 -10

4x=170

divide both sides by 4

x=42.5

6 0

2 years ago