Answer:
A. 
B. 
Step-by-step explanation:
Consider the equation
A. This equation has no solutions when the coefficients at x are the same and the free coefficients are not the same.
First, use distributive property:
So, the equation is
This equation has no solutions when
B. The equation has infinitely many solutions when the coefficients at x are the same and the free coefficients are the same too.
So, the equation
has infinitely many solutions when
In other cases, the equation has a unique solution
I think this is what you are looking for:
=−12
a
b
=
−
12
+=4
a
+
b
=
4
(+)2=42
(
a
+
b
)
2
=
4
2
2+2+2=16
a
2
+
b
2
+
2
a
b
=
16
∴2+2=16+2×12=40
∴
a
2
+
b
2
=
16
+
2
×
12
=
40
Now, (−)2=2+2−2=40+2×12=64
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
=
40
+
2
×
12
=
64
∴(−)=64‾‾‾√=±8
∴
(
a
−
b
)
=
64
=
±
8
So, 2−2=(+)(−)
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
2−2=(4)(±8)=±32
a
2
−
b
2
=
(
4
)
(
±
8
)
=
±
32
Hope that helps and sorry if it is confusing!
Answer:
1/3
Step-by-step explanation:
2/6=1/3
Answer:
Avoid histograms with small bin widths that group data into lots of bins. A histogram constructed with small bin widths will show the distribution as a “pancake.” so no it does not help us see the pattern in the data.
Step-by-step explanation:
