Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x
Crossing the 1s , the 2xs, and the x^2s out, we get
(2x</span>∆x+∆x^2-2∆x)/<span>∆x
Dividing the </span><span>∆x, we get
2x+</span><span>∆x-2.
Making the limit of </span><span>∆x=0, we get 2x-2.</span>
Answer:
<em>AAS</em>
Step-by-step explanation:
<em>because</em><em> </em><em>here </em><em>it </em><em>is </em><em>given</em><em> </em><em>that </em><em>two </em><em>angle </em><em>are</em>
<em> </em><em>equal</em>
and a side is common between both traingle
so, both traingle are congruent by
<em><u>AAS</u></em>
hope it helps
Range is ] -20 , -10 [
both points are disclude according to the empty circle
You need to subtract 25 by 18
Answer: The correct answer is “this is a reflection across the x-axis”
