You cannot solve for x but you can factor
to solve in ax^2+bx+c form you must find
b=x+y
a times c=x times y
so solve
3 times 3=9
what 2 number multily to get 9 and add to get 10
9=x times y
the numbers are 9 and 1 so
3x^2+1x+9x+3
(3x^2+1x)+(9x+3)
factor
(x)(3x+1)+(3)(3x+1)
reverse distribute
ab+ac=a(b+c)
(x)(3x+1)+(3)(3x+1)=(x+3)(3x+1)
factored out form is (x+3)(3x+1)
<span>(5h^3 − 8h) + (−2h^3 − h^2 − 2h)
= 5h^3 - 8h - 2h^3 - h^2 - 2h
= 3h^3 - h^2 - 10h</span>
1) to this case you must match both equations
3x+2 = 2x-3 ⇒3x-2x = -3-2⇒x= -5
now the value of x substitute it in the any of two equations, y = 3(-5)+2 = -13
and the other equation is the same value
y = 2(-5)-3 = -13
the point is ( -5,-13)
Before going to B.two-digit addition
By "density" I assume you mean "probability density function". For this to be the case for
, we require
Since
you have
which means
