You use the formula below...,
Angle theta = cos^-1 [scalar product of the two vectors ÷ multiplication of their moduli]
For example...the angle between
(2i + 3j) and (4i + 5j)
Angle theta = cos^-1 [( 2(4) + 3(5))÷((square root of (2^2 + 3^2))×(square root of (4^2 + 5^2))
Angle theta = cos^-1 [ 23 ÷ ((square root of(13)) × (square root of (41)]
Angle theta = 5°
Answer:
The first term is 14.
Step-by-step explanation:
The general formula for this kind of sequence (arithmetic) is a(n) = a(1) + (n-1)c, where c is the common difference. Here, we have a(6) = -1 = a(1) + (6-1)(-3), or -1 +15 = a(1). The first term is 14.
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:
D. Part of the solution region includes a negative number of erasers purchased; therefore, not all solutions are viable for the given situation.
Step-by-step explanation:
I got it right on the practice
