First, lets transform the given vector into an unit vector (dividing by its module)
UnitVec = 4/5 i + 3/5 j
Then lets change this vector into a polar form
UnitVec = 1. with angle of 36.869 degrees taking as a reference the i vector
Then, the probem tells us that the vectors u and v make an angle of 45 degrees with UnitVec, so lets add+-45 to the vector in polar form
U = 1*[cos(36.869 +45)i + sin(36.869 +45)j] = 0.1414 i + 0.9899 j
V = 1*[cos(36.869 -45)i + sin(36.869 -45)j] = 0.9899 i - 0.1414 j
Answer:
<h3>
= -1 </h3><h3>
*
= 1.4142</h3><h3>
= π =3.14159</h3><h3>
= π² = 9.8696</h3>
Explanation:
1.) Rewrite √24 = 
Rewrite √54 =
Divide both terms by the denominator; √6 cancels. 2-3 = -1
2.) Rewrite as 
×
The 3's cancel. 10/2 = 5√2 in the numerator.
√5 × √5 = 5 in the denominator. The 5's cancel.
That leaves √2 ≈1.4142
3.) Divide the terms in the numerator by the term in the denominator.
√2's cancel. 10π/2 = 5π 8π/2 = 4π
Subtract and we are left with π = 3.14159
4.) The square roots are reciprocals. They multiply to 1
We are left with π × π = π² ≈ 9.8696
<h3>
</h3>
Consecutive integers are 1 apart
x,x+1,x+2
(x)(x+1)(x+2)=-120
x^3+3x^2+3x=-120
add 120 to both sides
x^3+3x^2+3x+120=0
factor
(x+6)(x^2-3x+20)=0
set each to zero
x+6=0
x=-6
x^2-3x+20=0
will yeild non-real result, discard
x=-6
x+1=-5
x+2=-4
the numbers are -4,-5,-6
use trial and error and logic
factor 120
120=2*2*2*3*5
how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120?
obviously, the 5 has to stay since 2*5=10 which is out of range
so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7
obviously, 7 cannot happen since it is prime
3 and 4 results in in 12, but 2*2*2*3=24
therfor answer is 4 and 6
they are all negative since negaive cancel except 1
the numbers are -4,-5,-6
Y = 6x - 6 therefore y - 6x = -6
y - 6x = 5
so:
-6 ≠ 5
There is no solution to this system.
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