Ok to find dy/dx of x+2y=xy we take derivative of both sides with respect to x
1+2dy/dx = x*dy/dx +y*dx/dx
1+ 2dy/dx = x*dy/dx + y* 1
2dy/dx +1 = x*dy/dx + y
2y’ + 1 = xy’ + y
2y’ + 1 - xy’ = y
2y’ -xy’ = y - 1
y’(2-x) = y - 1
so we get finally
y’= (y-1)/(2-x)
Hope this helps you understand the concept! Any questions please ask! Thank you so much!!
<u>Part</u><u> </u><u>(</u><u>i</u><u>)</u>
1) AB is perpendicular to BC, ED is perpendicular to CD, BC = CD (given)
2) Angles ABC and CDE are right angles (perpendicular lines form right angles)
3) Angles ABC and CDE are equal (all right angles are equal)
4) Angles ACB and DCE are equal (vertical angles are equal)
5) Triangles ABC and EDC are congruent (ASA)
<u>Part</u><u> </u><u>(</u><u>ii</u><u>)</u>
6) AB = DE (corresponding parts of congruent triangles are equal)
Answer:
12.54
Step-by-step explanation:
5>0
Answer:
a number, t, increased by 23
a number, t, plus 23
Step-by-step explanation:
Have a lovely rest of your day/night, and good luck with your assignments! ♡
~ ren ⚘