$\text{The zeros:}\\\\x^3-x^2+9x-9=0\\\\x^2(x-1)+9(x-1)=0\\\\(x-1)(x^2+9)=0\iff x-1=0\ \vee\ x^2+9=0\\\\x-1=0\qquad\text{add 1 to both sides}\\x-1+1=0+1\\\boxed{x=1}\\\\x^2+9=0\qquad\text{subtract 9 from both sides}\\x^2=-9$

8 0

3 years ago

4) The triangle on the right is an equilateral <br><br> Find the value of X and Y

ozzi

Answer:

<em>angle Y is 29 degrees</em>

<em>angle X is 120 degrees</em>

Step-by-step explanation:

Well, if the triangle is an equilateral, then all the angles are equal to

60 degrees

that means the the angle at the top point is 60+31 = 91 degrees

the right point is 60 degrees

every triangle has 180 degrees total

180 - (91+60) = 29 degrees

NOW FORGET ABOUT THE EQUILATERAL FOR A BIT

180 = 31+x+29

simplify

180 = 60+x

subtract 60 from both sides

120 = x

7 0

2 years ago