Let's say make the shapes, some variable name
say
⛁ ⚪ ◽ ▯
a b c d
so.. if take those are the shapes provided
then we can take a look at the first set and we can say that
2a + b + c = 20
a + 4d + b = 19
a + c = 9
3c + d + a = 19
thus
and surely you can find the other two
A. that is b^2-4ac
for
ax^2+bx+c
the deterimant (part under the radicand)
is b^2-4ac
if it is
1. less than 0 then there are no real solutions
2. eqal to 0 then there is 1 real solution
3. greater than 0, there are 2 real roots
9x^2-16x+60
b^2-4ac=(-16)^2-4*9*60=256-2160<0
no real solutions
B. grouping
ac method
4 times -5=-20
what 2 numbers multiply to get -20 and add to get 8?
-2 and 10
seperate the middle number to get that
4x^2-2x+10x-5=0
(4x^2-2x)+(10x-5)=0
2x(2x-1)+2(2x-1)=0
(2x-1)(2x+2)=0
set equal to 0
2x-1=0
2x=1
x=1/2
2x+2=0
2x=-2
x=-1
x=1/2 or -1
19 over 4 is your answer unless you want me to simplify it for u
Answer:
-4
Step-by-step explanation:
its negative because it is decreasing from left to right and 4 because it goes down 4 over one not down one over four
Answer:134
Step-by-step explanation:
