Dy/dx = y/x^2
1/y dy = 1/x^2 dx
int(1/y dy) = int(1/x^2 dx)
ln y + c1 = -x + c2
ln y = -x + c ; c= c2-c1
y = e^(-x+c) = e^c e^-x
y = Ce^-x ; C=e^c
Answer:
1) 5 ( 4h - 1 )
2) 5 ( 3r + 7 )
3) 2( -2p +5 )
Step-by-step explanation:
For no. 1 you have to find a <u>common factor (</u>a number that divides into both 20 and 5) which is 5 because 20 divided by 5 is 4 and 5 divided by 5 is 1. so you have to do 20h divided by 5 which is 4h (don't forget the h ! ) and you do -5 divided by 1 which is -1. Remember when you do any multiplying or dividing questions with negatives and positives :
minus + minus = minus (-5*-6 = -30)
positive + positive = positive (5*6 = 30) * equals multiplacation by the way
minus + positive = minus (-5*6 = -30)
positive + minus = minus (5*-6 = -30)
hope this helped xx
(x–4) (3x^2 + 2)
i hope that helps with you’re expression
