(x - 2)²(x+1)
(x-2)(x-2)
multiply the two brackets together
(x)(x)=x^2
(x)(-2)=-2x
(-2)(x)=-2x
(-2)(-2)=4
x^2-2x-2x+4
x^2-4x+4
(x^2-4x+4)(x+1)
multiply the brackets together
(x^2)(x)=x^3
(x^2)(1)=x^2
(-4x)(x)=-4x^2
(-4x)(1)=-4x
(4)(1)=4
x^3+x^2-4x^2-4x+4x+4
Answer:
x^3-3x^2+4
Answer:
The answer is b.
Step-by-step explanation:
Answer:
Step-by-step explanation:
hello :
16t²+120 = y
16t² = y -120 so : t² = (y-120)/16
if : y-120 ≥ 0 t = ±√((y-120)/16)
Looks like all you have to do it multiply the numbers together; there is no application of the Distributive Property to these problems.
First picture)
I: 5x+2y=-4
II: -3x+2y=12
add I+(-1*II):
5x+2y-(-3x+2y)=-4-12
8x=-16
x=-2
insert x=-2 into I:
5*(-2)+2y=-4
-10+2y=-4
2y=6
y=3
(-2,3)
question 6)
I: totalcost=115=3*childs+5*adults
II: 33=adults+childs
33-adults=childs
insert childs into I:
115=3*(33-adults)+5*adults
115=99-3*adults+5*adults
16=2*adults
8=adults
insert adults into II:
33-8=childs
25=childs
so it's the last option
question 7)
a) y<6 and y>2 can also be written as 2<y<6, so solution 3 exist for example
b) y>6 and y>2 can also be written as 2<6<y, so solution 7 exist for example
c) y<6 and y<2 inverse of b: y<2<6, so for example 1
d) y>6 and y<2: y<2<6<y, this is impossible as y can be only either bigger or smaller than 2 or 6
so it's the last option
question 8)
I: x+y=12
II: x-y=6
subtract: I-II:
x+y-(x-y)=12-6
2y=6
y=3
insert y into I:
x+3=12
x=9
(9,3)
question 9)
I: x+y=6
II: x=y+5
if you take the x=y+5 definition of II and substitute it into I:
(y+5)+y=6
which is the second option :)