We can not see the rest of 12
Answer:
The answer is one solution because the graphs are together in one specific point.
Answer:
48f^2-110f+63
Step-by-step explanation:
We multiple every part of first sum with every prt of the 2nd sum:
(6f-7)(8f-9)=48f^2-54f-56f+63=
48f^2-110f+63
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=

so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=

x=

x=

x=

x=

there are no real numbers that satisfy this